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Numbers in CAT CMAT and other exams
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Scope and syllabus for Numbers in CAT, CMAT, NMAT, SNAP and other entrance exams:

• Number Tree
• LCM HCF
• Divisibility Rules
• Power cycle
• Remainder Theorem
• Last and Second last digit
• Remainder of powers an – bn
• Power of Exponents
• Euler’s Theorem
• Fermet’s Theory
• Number Systems (decimal binary)
Importance in Numbers in CAT, CMAT, NMAT, SNAP and other entrance exams
Numbers in CAT – 4 – 5 questions
Numbers in CMAT – 2 – 3 questions
Numbers in NMAT – 5 questions
Numbers in SNAP – 2 ques

Ex.1. which of the following are prime numbers? (i) 241 (ii) 337 (iii) 391

A) 241,337 B) 337,391 C) all the above D) none

Sol. (i) Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, and 13. 241 is not

Divisible by any one of them. 241 is a prime number.

(ii)Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, and 17. 337 is not

Divisible by any one of them. 337 is a prime number.

(iii)Clearly, 20 > Ö39l”. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19.

We find that 391 is divisible by 17. 391 is not prime.

Ex.2. Simplify: (i) 896 x 896 – 204 x 204 (ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

A) 761000,251000 B) 761200,251001 C) 761200,251000 D) 76100, 25100

Sol. (i) Given exp = (896)2 – (204)2 = (896 + 204) (896 – 204) = 1100 x 692 = 761200.

(ii) Given exp = (387)2+ (114)2+ (2 x 387x 114)

= a2 + b2 + 2ab, where a = 387, b=114

Numbers and Fractions

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= (a+b) 2 = (387 + 114)2 = (501)2 = 251001.

Ex.3. which of the following numbers is divisible by 3? (i) 541326 (ii) 5967013

A) i & ii B)only (ii) C)only (i) D)none

Sol.

(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.Hence, 541326

is divisible by 3.

(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.

Hence, 5967013 is not divisible by 3.

Ex.4. which one of the following is not a prime number?

A)   31 B)61 C)71 D)91

Sol: 91 is divisible by 7.so, it is a prime number

Ex.5.What least value must be assigned to * so that the number 197*5462 is r 9 ?

A)1 B)0 C)2 D)3

Sol: Let the missing digit be x.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +2) = (34 + x).

For (34 + x) to be divisible by 9, x must be replaced by 2 .

Hence, the digit in place of * must be 2.

Ex.6. Which of the following numbers is divisible by 4 ? (i) 6792059 (ii) 618703572

A) i & ii B)only (ii) C)only (i) D)none (SBI PO 2000)

Sol. (i) The number formed by the last two digits in the given number is 94, which is not

Numbers and Fractions

14

divisible by 4. Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible

by 4. Hence, 618703572 is divisible by 4.

Ex.7. which digits should come in place of * and \$ if the number 62684*\$ is divisible by

both 8 and 5?

A)   4,0 B)0,4 C)4,4 D)0,0

Sol. Since the given number is divisible by 5, so 0 or 5 must come in place of \$. But, a number

ending with 5 is never divisible by 8. So, 0 will replace \$. Now, the number formed by the last

three digits is 4*0, which becomes divisible by 8, if * is replaced by 4. Hence, digits in place of *

and \$ are 4 and 0 respectively.

Ex.8. Is 4832718 is divisible by 11.

A)   yes B)no C)none

Sol.(Sum of digits at odd places) – (Sum of digits at even places) = (8 + 7 + 3 + 4) – (1 + 2 + 8)

= 11, which is divisible by 11. Hence, 4832718 is divisible by 11.

Ex.9. what least number must be added to 3000 to obtain a number exactly divisible by

19?

A) 5 B) 3 C) 2 D) 0

Sol. On dividing 3000 by 19, we get 17 as remainder.

Number to be added = (19 – 17) = 2.

Ex.10. what least number must be subtracted from 2000 to get a number exactly divisible

by 17?

A) 11 B) 2 C) 5 D) 1

Sol. On dividing 2000 by 17, we get 11 as remainder. Required number to be subtracted = 11.

Ex.11. Find the smallest number of 6 digits which is exactly divisible by 111.

A)   100000 B)100011 C)100111 D)100110

Sol. Smallest number of 6 digits is 100000.On dividing 100000 by 111, we get 100 as remainder.

Number to be added = (111 – 100) – 11. Hence, required number = 100011.

Ex.12. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37.

Find the divisor.

A)   180 B)179 C)184 D)25

Dividend – Remainder        15968-37

Sol. = ————————– = ————- = 179.

.Quotient      Divisor                        89

Ex.13. A number when divided by 342 gives a remainder 47. When the same number is

divided by 19, what would be the remainder?

A) 5 B) 9 C) 4 D) 0

Sol. On dividing the given number by 342, let k be the quotient and 47 as remainder.

Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.

The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex.14. Find the remainder when 231 is divided by 5.

A)   4 B)5 C)3 D)7

Sol: 210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].

Unit digit of 231 is 8. Now, 8 when divided by 5, gives 3 as remainder.

Hence, 231 when divided by 5, gives 3 as remainder.

Ex.15. Find the sum of all odd numbers up to 100.

A)   2000 B)2500 C)2800 D)3000

Sol. The given numbers are 1, 3, 5, 7, …, 99. This is an A.P. with a = 1 and d = 2.

Let it contain n terms. Then, 1 + (n – 1) x 2 = 99 or n = 50.

Required sum = n (first term + last term)

2

= 50 (1 + 99) = 2500.

2

Ex.16. Find the sum of all 2 digit numbers divisible by 3.

A)   1700 B)1665 C)1600 D)1605

Sol. All 2 digit numbers divisible by 3 are: 12, 51, 18, 21… 99. This is an A.P. with a = 12 and

d = 3. Let it contain n terms. Then,12 + (n – 1) x 3 = 99 or n = 30.

Required sum = 30 x (12+99) = 1665.

2

Ex.17.It is being given that (232 + 1) is completely divisible by a whole number. Which of

the following numbers is completely divisible by this number?

A)(216+1) B)(216-1) C)(7*223) D)(296+1)

Sol: Let 232 = x. Then, (232 + 1) = (x + 1). Let (x + 1) be completely divisible by the natural

number N. Then, (296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 – x + 1), which is completely

divisible by N, since (x + 1) is divisible by N.

Ex.18. How many prime numbers are less than 50?

A)   16 B)15 C)14 D)18

Sol: Prime numbers less than 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Their number is 15

Ex.19.Which natural number is nearest to 9217, which is completely divisible by 88 ?

A)   9152 B)9240 C)9064 D)9184

Sol: On dividing we get,

88) 9217 (104

88

417

352

65

Therefore, required number = 9217 + (88 -65) = 9217 + 23 = 9240.

Ex.20: 3251 + 587 + 369 – ? = 3007

A) 1250 B) 1300 C) 1375 D) 1200

Sol: 3251 Let 4207 – x = 3007

+ 587 Then, x = 4207 – 3007 = 1200

+ 369

4207

Ex.21. If the product 4864 x 9 P 2 is divisible by 12, then the value of P is:

A) 2 B) 5 C) 6 D) none

Sol: Clearly, 4864 is divisible by 4. So, 9P2 must be divisible by 3. So, (9 + P + 2) must be

divisible by 3. P = 1.

Ex.22. In a division sum, the divisor is 10 times the quotient and 5 times the remainder. If

the remainder is 46, what is the dividend ?

A)   4236 B)4306 C)4336 D)5336

Sol: Divisor = (5 x 46) = 230 10 x Quotient = 230

230

= 23

10

Dividend = (Divisor x Quotient) + Remainder = (230 x 23) + 46 = 5290 + 46 = 5336.