Ans .
18! contains 15 and 5, which combined with one even number give zeroes. Also, 10 is also contained in 18!, which will give an additional zero. Hence, 18! contains 3 zeroes and the last digit will always be zero.
Ans .
6
Number of zeroes is 27! = [27/5] + [27/25] where [x] indicates the integer just lower than the fraction Hence, [27/5] = 5 and [27/25] = 1, Hence 6 zeroes
Ans .
n [137/5] + [137/52] + [137/53] = 27 + 5 + 1 = 33 zeroes
Ans .
7
since 8 is not prime, use the following process. The prime factors of 8 is 2 × 2 × 2. So we do [25/2] + [25/22] + [25/23] + [25/24] + [25/25] = 12 + 6 + 3 + 1 = 22. However we need highest power of 8 which is 23. So we divide 22/3 to get 7 and an additional 2. since 25! has 22 twos, it will be divided by 8 seven times.
Ans .
8
The question can be converted to the form 24n+3 * 74n which is given as 8 * 1 = 8 . hence last digit is 8
Ans .
101,100,99
(i) Both ends included-Solution: 200 – 100 + 1 = 101 (ii) One end included-Solution: 200 – 100 = 100 (iii) Both ends excluded-Solution: 200 – 100 – 1 = 99.
Ans .
3,2850
114 = 19 * 3 * 2; 75 = 52 * 3; GCD = 3 ; LCM = 150*19 = 2850
Ans .
61,60,59
(i) Both ends included—Solution: (242 – 122)/2 = 60 + 1 = 61 (ii) One end included-Solution: (242 – 122)/2 = 60 (iii) Both ends excluded-Solution: (242 – 122)/2 – 1 = 59
Ans .
8
The number of zeroes depends on the number of fives and the number of twos. Here, close scrutiny shows that the number of twos is the constraint. The expression can be written as 5 * (5 * 2) * (5 * 3) * (5 * 2 * 2) * (5 * 5) * (5 * 2 * 3) * (5 * 7) * (5 * 2 * 2 * 2) * (5 * 3 * 3) * (5 * 5 * 2) Number of 5s – 12, Number of 2s – 8. Hence: 8 zeroes
Ans .
1
The remainder for the expression: [(73 * 79 * 81)/11] will be the same as the remainder for [(7 * 2 * 4)/11] That is, 56/11 so remainder = 1
Ans .
1
(3560/8) = [(32)280/8] = (9280/8) = [9.9.9…(280 times)]/8. remainder for above expression = remainder for [1.1.1…(280 times)]/8 , remainder = 1.
Ans .
0
2222 divided by 7 gives 3. So we have 35555/7, which we can write as 3*(32)5555 / 7 , which becomes 3*25555 / 7 = 3*22*8925 / 7 = Remainder is 5 since 3*4*1/7 = 12/7 = 5. For the second number we write 42222/7 = 2 * 81481 / 7 = Remainder is 2. Now the euation is [2 + 5]/7 which gives remainder as 0.
Ans .
126 = 2 * 7 * 32 ; 540 = 22 * 33 * 5 ; 630 = 2 * 5 * 7 * 32; GCD = 2 * 32 = 18; LCM = 22 * 33 * 5 * 7 = 3780;
Ans .
8
Since B is the largest digit, option (a) is rejected. Check for option (b). If B is 6, then the two largest two-digit numbers are 65 and 60 (Since, their difference is 5) and we have B = 6, A = 5 and C = 0. But with this solution we are unable to meet the second condition. Hence (b) is not the answer. We also realise here that C cannot be 0. Check for option (c). B is 7, then the nos. are 76 and 71 or 75 and 70. In both these cases, the smallest two two-digit numbers do not differ by 2. Hence, the answer is not (c). Hence, option (d) is the answer
Ans .
18
n We use the remainder theorem to solve the problem. Using the theorem, we see that the following expressions have the same remainder. Thus, [22 * 10 * 10 * 21 * 21 * 21] / 23 = [-1*-2*-2*-2*100]/23 = 8 * 100 / 23 = 18
Ans .
6
We can rewrite the expression as 84n+2 * 64n * 74n+2 = 4 * 6 * 9 = 6
Ans .
39
The answer will be given by the HCF of 378 and 675. HCF is 27. number of sections are 378/27 + 675/27 = 14+25 = 39
Ans .
99 × difference between X and Z
From the property of numbers, it is known that on reversing a three digit number, the difference (of both the numbers) will be divisible by 99. Also, it is known that this difference will be equal to 99 × difference between the units and hundreds digits of the three digit number. fi Option (d)
Ans .
0
we can say that the difference will be divisible by 99 . Remainder = 0
Ans .
4
The quotient will be the difference between extreme digits of 783, i.e. 7 – 3 = 4
Ans .
10
As 242 is divisible by 22, so the required length of left wood will be equal to the remainder when 98 is divided by 22: Hence, 10 [98/22; remainder 10]
Ans .
1
when you see a remainder of – 4 when the number is divided by 5, the required remainder will be equal to 5 – 4 = 1.
Ans .
5.2
When -24.8 is divided by 6 we get remainder -0.8 but we cannot have negative remainder so 6-0.8=5.2
Ans .
q – 1
Minimum remainder is 0. Maximum possible remainder = q – 1 So, required maximum possible difference = (q – 1) – 0 = (q – 1).
Ans .
1
2256/17 = 1664/17 = 1
Ans .
0
784 / 342 = (73)28 / 342 = 1; 784 / 344 = (73)28 / 344 = 1. 1 - 1 = 0
Ans .
8
(10017 - 1) = 10....00 (34 zeroes) - 1 = 99..99 / 9 (Remainder is 0). (1034 + x) = 10....0 + x (34 zeroes) OR 1000..0x (33 zeroes). This to be divided by x has to be 8.
Ans .
0
The units digit in this case would obviously be ‘0’ because the given expression has a pair of 2 and 5 in it’s prime factors.
Ans .
The difference between the digits would be 54/9 = 6. Hence only option a is correct
Ans .
The two numbers should be factors of 405. A factor search will yield the factors. (look only for 2 digit factors of 405 with sum of digits between 1 to 19). Also 405 = 5 × 34. Hence: 15 × 27, 45 × 9 are the only two options. From these factors pairs only the second pair gives us the desired result. i.e. Number × sum of digits = 405. Hence, the answer is 45
Ans .
12, 3
It can be seen that Option (b) 12,3 fits the situation perfectly as their Arithmetic mean = 7.5 and their geometric mean = 6 and the geometric mean is 20% less than the arithmetic mean
Ans .
49, 1
Two more than half of 1/3 rd of 96 = 18. Also since we are given that the difference between the AM and GM is 18, it means that the GM must be an integer. From amongst the options, only option (a) gives us a GM which is an integer
Ans .
7
To be divisible by 11, A + 8 - (3+1) = A+4 be a multiple of 11 or be 0. A = 7.
Ans .
6
For 381A to be divisible by 9, the sum of the digits 3 + 8 + 1 + A should be divisible by 9. For that to happen A should be 6. Option (d) is correct.
Ans .
2
The remainder of 96 / 8 is 1 and when 1 more is added then we get remainder 2
Ans .
12:1
LCM of 5, 15 and 20 = 60. HCF of 5, 15 and 20 = 5. The required ratio is 60:5 = 12:1
Ans .
440
LCM of 5/2, 8/9 and 11/14 would be given by: (LCM of numerators)/(HCF of denominators) = 440/1 = 440
Ans .
3A will always be divisible by 6
Only the first option can be verified to be true in this case. If A is even, 3A would always be divisible by 6 as it would be divisible by both 2 and 3. Options b and c can be seen to be incorrect by assuming the value of A as 4.
Ans .
4
The essence of this question is in the fact that the last digit of the number is 0. Naturally, the number is necessarily divisible by 2,5 and 10. Only 4 does not necessarily divide it
Ans .
Both (a) and (c)
B would necessarily be even- as the possible values of B for the three digit number 15B to be divisible by 6 are 0 and 6. Also, the condition stated in option (c) is also seen to be true in this case – as both 0 and 6 are divisible by 6
Ans .
0
The units digit would be given by 5 + 6 + 9 (numbers ending in 5 and 6 would always end in 5 and 6 irrespective of the power and 354 will give a units digit equivalent to 34n+2 which would give us a unit digit of 9)
Ans .
270
The number of zeroes would be given by adding the quotients when we successively divide 1090 by 5: 1090/5 + 218/5 + 43/5 + 8/5 = 218 + 43 + 8 + 1 = 270
Ans .
35
The number of 5’s in 146! can be got by [146/5] + [29/5] + [5/5]= 29+5+1 = 35
Ans .
12
1420 = 142 × 10 = 22 × 71 × 5. Thus, the number of factors of the number would be (2 + 1) (1 + 1) (1 + 1) = 3 × 2 × 2 = 12.
Ans .
(x - 2), (x – 2) (x – 3) (x – 5)
= (x – 2)(x – 3) & (x 2 – 7x + 10) = (x – 5)(x – 2) Required HCF = (x – 2); required LCM = (x – 2)(x – 3)(x – 5)
Ans .
4
Since both P and Q are prime numbers, the number of factors would be (1 + 1)(1 + 1) = 4
Ans .
6
Since both P and Q are prime numbers, the number of factors would be (2 + 1)(1 + 1) = 6
Ans .
12
Since both P and Q are prime numbers, the number of factors would be (3 + 1)(2 + 1) = 12
Ans .
9
The sides of the pentagon being 1422, 1737, 2160, 2214 and 2358, the least difference between any two numbers is 54. Hence, the correct answer will be a factor of 54. Further, since there are some odd numbers in the list, the answer should be an odd factor of 54. Hence, check with 27, 9 and 3 in that order. You will get 9 as the HCF
Ans .
9793
The LCM of the 4 numbers is 612. The highest 4 digit number which would be a common multiple of all these 4 numbers is 9792. Hence, the correct answer is 9793.
Ans .
2884
The LCM of 16, 18 and 20 is 720. The numbers which would give a remainder of 4, when divided by 16, 18 and 20 would be given by the series: 724, 1444, 2164, 2884 and so on. Checking each of these numbers for divisibility by 7, it can be seen that 2884 is the least number in the series that is divisible by 7 and hence is the correct answer
Ans .
72
They will ring together again after a time which would be the LCM of 6, 8, 12 and 18. The required LCM = 72. Hence, they would ring together after 72 seconds
Ans .
10
720/72 = 10 times
Ans .
0
5 × 7 × 6 = 0
Ans .
none
All these numbers can be verified to not be perfect squares
Ans .
x 000 where x is a natural number
A perfect square can never end in an odd number of zeroes
Ans .
9
It is obvious that the LCM of 5,12,18 and 20 would never be a multiple of 9. At the same time it has to be a multiple of each of 3, 8 and 5.
Ans .
30
720 = 24 × 32 × 51. Number of factors = 5 × 3 × 2 = 30
Ans .
None of these
16 – x2 = (4 – x) (4 + x) and x2 + x – 6 = (x + 3)(x – 2) The required LCM = (4 – x)(4 + x) (x + 3) (x – 2).
Ans .
x – 2
x2 – 4 = (x – 2) (x + 2) and x2 + x – 6 = (x + 3) (x - 2) GCD or HCF of these expressions = (x – 2)
Ans .
2 × A
If A is not divisible by 3, it is obvious that 2A would also not be divisible by 3, as 2A would have no ‘3’ in it.
Ans .
1
Since this is of the form (a + 1)n/a, the Remainder = 1
Ans .
1
(a)EVEN POWER/(a + 1). The remainder = 1 in this case as the power is even
Ans .
10,10
The condition for the product to be the greatest is if the two terms are equal. Thus, the break up in option (a) would give us the highest product of the two parts.
Ans .
12
50/5 =10, 10/5 =2. Thus, the required answer would be 10 + 2 = 12.
Ans .
13,62,480
to be divisible by 24, number must be divisible by3,8 both
Ans .
All of these
Any number divisible by 88, has to be necessarily divisible by 11, 2, 4, 8, 44 and 22. Thus, each of the first three options is correct
Ans .
60
10800 = 108 × 100 = 33 × 24 × 52 .The number of divisors would be: (3 + 1) (4 + 1) (2 + 1) = 4 × 5 × 3 = 60 divisors. Option (b) is correct.
Ans .
all
Three consecutive natural numbers, starting with an even number would always have at least three 2’s as their prime factors and also would have at least one multiple of 3 in them. Thus, 6, 12 and 24 would each divide the product.
Ans .
3
When the birds sat one on a branch, there was one extra bird. When they sat 2 to a branch one branch was extra. To find the number of branches, go through options. Checking option (a), if there were 3 branches, there would be 4 birds. (this would leave one bird without branch as per the question.) When 4 birds would sit 2 to a branch there would be 1 branch free (as per the question)
Ans .
1 always
The number would either be (3n + 1)2 or (3n + 2)2. In the expansion of each of these the only term which would not be divisible by 3 would be the square of 1 and 2 respectively. When divided by 3, both of these give 1 as remainder
Ans .
15
get the prime factors and reduce the expression
Ans .
-7
The upper limit for x + y = 4 + 3 = 7. The lower limit of x – y = 2 – 3 = –1. Required ratio = 7/–1 = –7.
Ans .
32
For the sum of squares of digits to be 13, it is obvious that the digits should be 2 and 3. So the number can only be 23 or 32. Further, the number being referred to has to be 32 since the reduction of 9, reverses the digits
Ans .
54 and 45
trying the value in the options you get that the product of 54 × 45 = 2430.
Ans .
90 and 24
Option (b) can be verified to be true as the LCM of 90 and 24 is indeed 360
Ans .
78 and 13 or 26 and 39
The pairs given in option (d) 78 and 13 and 26 and 39 meet both the conditions of LCM of 78 and HCF of 13. Option (d) is correct
Ans .
51 and 34
Solve using options. Option (d) 51 and 34 satisfies the required conditions.
Ans .
all
2,4,6,8,0 at the end makes number divisible by 2
Ans .
3
the last digit being 1 can never be obtained if x is 2,5,6.
Ans .
Exactly 11
Ties < Trousers < Shirts. Since each of the three is minimum 11, the total would be a minimum of 33 (for all 3). The remaining 5 need to be distributed amongst ties, trousers and shirts so that they can maintain the inequality Ties < Trousers < Shirts This can be achieved with 11 ties, and the remaining 27 pieces of clothing distributed between trousers and shirts such that the shirts are greater than the trousers. This can be done in at least 2 ways: 12 trousers and 15 shirts; 13 trousers and 14 shirts. If you try to go for 12 ties, the remaining 26 pieces of clothing need to be distributed amongst shirts and trousers such that the shirts are greater than the trousers and both are greater than 12. With only 26 pieces of clothing to be distributed between shirts and trousers this is not possible. Hence, the number of ties has to be exactly 11. Option (a) is correct.
Ans .
At least 14
The number of shirts would be at least 14 as the two distributions possible are: 11, 12, 15 and 11, 13, 14. Option (b) is correct.
Ans .
184
The LCM of 12, 15, 18 and 20 is 180. Thus, the least number would be 184. Option (c) is correct.
Ans .
7
2800 = 20 × 20 × 7. Thus, we need to multiply or divide with 7 in order to make it a perfect square.
Ans .
1024
square of 32 is 1024
Ans .
364
First find the LCM of 6, 9, 15 and 18. Their LCM = 18 × 5 = 90. The series of numbers which would leave a remainder of 4 when divided by 6, 9, 15 and 18 would be given by: LCM + 4; 2 × LCM + 4; 3 × LCM + 4; 4 × LCM + 4; 5 × LCM + 4 and so on . Thus, this series would be: 94, 184, 274, 364, 454…. The other constraint in the problem is to find a number which also has the property of being divisible by 7. Checking each of the numbers in the series above for their divisibility by 7, we see that 364 is the least value which is also divisible by 7
Ans .
121
LCM of 2, 3, 4, 5 and 6 = 6 × 5 × 2 =60 (Refer to the shortcut process for LCM given in the chapter notes). Thus, the series 61, 121, 181 etc would give us a remainder 1 when divided by 2, 3, 4, 5 and 6. The least 3 digit number in this series is 121
Ans .
35
70 = 2 × 5 × 7; 245 = 5 × 7 × 7. HCF = 5 × 7 = 35.
Ans .
91
7056 is the closest perfect square below 7147. Hence, 7147 – 7056 = 91 is the required answer.
Ans .
3600
The LCM of 6, 8 and 15 = 120. Thus, we need to look for a perfect square in the series of multiples of 120. 120, 240, 360, 480, 600, 720…., the first number which is a perfect square is: 3600.
Ans .
7
30492 = 22 × 32 × 71 × 112 For a number to be a perfect square each of the prime factors in the standard form of the number needs to be raised to an even power. Thus, we need to multiply or divide the number by 7 so that we either make it: 22 × 32 × 72 × 112 (if we multiply the number by 7) or We make it: 22 × 32 × 112 (if we divide the number by 7).
Ans .
9944
88 × 113 = 9944 is the greatest 4 digit number exactly divisible by 88
Ans .
51
3/4 th of 116 = ¾ × 116 = 87 4/5th of 45 = 4/5 × 45 = 36. Required difference = 51.
Ans .
75
The correct arrangement would be 75 plants in a row and 75 rows since 5625 is the square of 75
Ans .
7
9EVEN POWER × 74n+1 = 1 × 7 = 7 as the units digit of the multiplication.
Ans .
40,80
It can be seen that for 40 and 80 the number of factors are 8 and 10 respectively. Thus option (c) satisfies the condition.
Ans .
3988
In order to solve this question you need to realize that remainders of 1, 3, 4 and 5 in the case of 3, 5, 6 and 7 respectively, means remainders of –2 in each case. In order to find the number which leaves remainder –2 when divided by these numbers you need to first find the LCM of 3, 5, 6 and 7 and subtract 2 from them. Since the LCM is 210, the first such number which satisfies this condition is 208. However, the question has asked us to find the largest such number below 4000. So you need to look at multiples of the LCM and subtract 2. The required number is 3990 – 2 = 3988
Ans .
13
The number would be given by the (LCM of 2, 3 and 4) +1 Æ which is 12 + 1 = 13.
Ans .
210
They would ring together again after a time interval which would be the LCM of 5, 6 and 7. Since the LCM is 210,
Ans .
2
Since the numbers have their HCF as 16,both the numbers have to be multiples of 16 (i.e. 24). 7168 = 210 × 71 In order to visualise the required possible pairs of numbers we need to look at the prime factors of 7168 in the following fashion: 7168 = 210 × 71 = (24 × 24) × 22 × 71 = (16 × 16) × 2 × 2 × 7 It is then a matter of distributing the 2 extra twos and the 1 extra seven in 22 × 71 between the two numbers given by 16 and 16 inside the bracket. The possible pairs are: 32 × 224; 64 × 112; 16 × 448. Thus there are 3 distinct pairs of numbers which are multiples of 16 and whose product is 7168. However, out of these the pair 32 × 224 has it’s HCF as 32 and hence does not satisfy the given conditions. Thus there are two pairs of numbers that would satisfy the condition that their HCF is 16 and their product is 7168.