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Page No 5.20:

Question 1:

Given that the number 35α64¯ is divisible by 3, where α is a digit, what are the possible values of α?

Answer:

It is given that 35a64¯ is a multiple of 3. (3+5+a+6+4) is a multiple of 3. (a+18) is a multiple of 3. (a+18)=0, 3, 6, 9, 12, 15, 18, 21...But a is a digit of number  35a64 ¯ . So, a can take value 0, 1, 2, 3, 4...9.a+18=18 a=0a+18=21 a=3a+18=24 a=6a+18=27 a=9 a=0, 3, 6, 9

Page No 5.20:

Question 2:

If x is a digit such that the number 18x71¯ is divisible by 3, find possible values of x.

Answer:

It is given that 18x71¯ is a multiple of 3.(1+8+x+7+1) is a multiple of 3.(17+x) is a multiple of 3.17+x=0, 3, 6, 9, 12, 15, 18, 21...But x is a digit. So, x can take values 0,1,2,3,4...9.17+x=18 x=117+x=21 x=417+x=24 x=7x=1,4,7

Page No 5.20:

Question 3:

If x is a digit of the number 66784x¯ such that it is divisible by 9, find possible values of x.

Answer:

It is given that  66784x¯ is a multiple of 9.Therefore, (6+6+7+8+4+x) is a multiple of 9.And,(31+x) is a multiple of 9. Possible values of (31+x) are 0, 9, 18, 27, 36, 45,...But x is a digit. So, x can only take value 0, 1, 2, 3, 4,...9.31+x=36 x=36-31x=5

Page No 5.20:

Question 4:

Given that the number 67y19¯ is divisible by 9, where y is a digit, what are the possible values of y?

Answer:

It is given that 67y19¯ is a multiple of 9. (6+7+y+1+9) is a multiple of 9. (23+y) is a multiple of 9. 23+y=0, 9, 18, 27, 36...But x is a digit. So, x can take values 0,1,2,3,4...9.23+y=27y=4

Page No 5.20:

Question 5:

If 3x2¯ is a multiple of 11, where x is a digit, what is the value of x?

Answer:

Sum of the digits at odd places = 3+2= 5Sum of the digit at even place = x Sum of the digit at even place-Sum of the digits at odd places=(x-5) (x-5) must be multiple by 11. Possible values of (x-5) are 0, 11, 22, 33...But x is a digit; therefore x must be  0, 1 ,2, 3...9. x-5=0 x=5

Page No 5.20:

Question 6:

If 98215x2¯ is a number with x as its tens digit such that is is divisible by 4. Find all possible values of x.

Answer:

A natural number is divisible by 4 if the number formed by its digits in units and tens places is divisible by 4. 98215x2 will be divisible by 4 if x2 is divisible by 4.  x2 ¯=10x + 2x is a digit; therefore possible values of x are 0, 1, 2, 3...9.x2 ¯=2, 12, 22, 32, 42, 52, 62, 72, 82, 92The numbers that are divisible by 4 are 12, 32, 52, 72, 92.Therefore, the values of x are 1, 3, 5, 7, 9.

Page No 5.20:

Question 7:

If x denotes the digit at hundreds place of the number 67x19¯ such that the number is divisible by 11. Find all possible values of x.

Answer:

A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.Sum of digits at odd places-Sum of digits at even places=(6 + x + 9)-(7 + 1)=(15 + x)-8=x + 7 x + 7 =11  x = 4

Page No 5.20:

Question 8:

Find the remainder when 981547 is divided by 5. Do this without doing actual division.

Answer:

If a natural number is divided by 5, it has the same remainder when its unit digit is divided by 5.Here, the unit digit of 981547 is 7. When 7 is divided by 5, remainder is 2.Therefore, remainder will be 2 when 981547 is divided by 5.

Page No 5.20:

Question 9:

Find the remainder when 51439786 is divided by 3. Do this without performing actual division.

Answer:

Sum of the digits of the number 51439786 = 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43The remainder of 51439786, when divided by 3, is the same as the remainder when the sum of the digits is divided 3.When 43 is divided by 3, remainder is 1.Therefore, when 51439786 is divided by 3, remainder will be 1.

Page No 5.20:

Question 10:

Find the remainder, without performing actual division, when 798 is divided by 11.

Answer:

798 =A multiple of 11 + (Sum of its digits at odd places - Sum of its digits at even places)798 =A multiple of 11 + (7 + 8 - 9)798 =A multiple of 11 + (15 - 9)798 =A multiple of 11 + 6Therefore, the remainder is 6.

Page No 5.20:

Question 11:

Without performing actual division, find the remainder when 928174653 is divided by 11.

Answer:

928174653= A multiple of 11 + (Sum of its digits at odd places -Sum of its digits at even places)928174653= A multiple of 11 + {(9 + 8 + 7 + 6 + 3) - (2 + 1 + 4 + 5)}928174653= A multiple of 11 + (33 - 12)928174653= A multiple of 11 + 21928174653= A multiple of 11 + (11 × 1 + 10)928174653= A multiple of 11 + 10Therefore, the remainder is 10.

Page No 5.20:

Question 12:

Given an example of a number which is divisible by
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not by 32.

Answer:

(i) 10
Every number with the structure (4n + 2) is an example of a number that is divisible by 2 but not by 4.

(ii) 15
Every number with the structure (6n + 3) is an example of a number that is divisible by 3 but not by 6.

(iii) 28
Every number with the structure (8n + 4) is an example of a number that is divisible by 4 but not by 8.

(iv) 8
Every number with the  structure (32n + 8), (32n + 16) or (32n + 24) is an example of a number that is divisible by 4 and 8 but not by 32.

Page No 5.20:

Question 13:

Which of the following statements are true?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-prime, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.

Answer:

(i) False
Every number with the structures (9n + 3) or (9n + 6) is divisible by 3 but not by 9. Example: 3, 6, 12 etc.
(ii) True
(iii) False
Every number with the structure (8n + 4) is divisible by 4 but not by 8. Example: 4, 12, 20 etc.
(iv) True
(v) False
Example: 24 is divisible by both 3 and 6 but it is not divisible by 18.
(vi) True
(vii) False
Example: 5 divides 10, which is a sum of 3 and 7. However, it neither divides 3 nor 7.
(viii) True
(ix) False
Example: 4 and 9 are co-prime numbers but both are composite numbers too.
(x) True



Page No 5.30:

Question 1:

Solve each of the following Cryptarithms:
37+AB 9A

Answer:

Two possible values of A are:(i) If 7 + B  9 3 + A = 9  A = 6But if A = 6, 7 + B must be larger than 9. Hence, it is impossible.(ii) If 7 + B  9 1 + 3 + A = 9 A = 5If A = 5 and 7 + B = 5, B must be 8 A = 5, B = 8

Page No 5.30:

Question 2:

Solve each of the following Cryptarithm:
    A   B+3   7    9   A¯

Answer:

Two possibilities of A are:(i) If B + 7  9, A = 6But clearly, if A = 6, B + 7  9; it is impossible(ii) If B + 7  9, A = 5 and B + 7 = 5 Clearly, B = 8 A = 5, B = 8

Page No 5.30:

Question 3:

Solve each of the following Cryptarithm:
A1+1B B0

Answer:

If 1 + B = 0   Surely, B = 9If 1 + A + 1 = 9  Surely, A = 7

Page No 5.30:

Question 4:

Solve each of the following Cryptarithm:
2AB+AB1 B18¯

Answer:

B + 1 = 8, B = 7A + B = 1, A + 7 = 1, A = 4So, A = 4, B = 7

Page No 5.30:

Question 5:

Solve each of the following Cryptarithm:
12A+6AB A09¯

Answer:

A + B = 9  as the sum of two digits can never be 192 + A = 0, A must be 8A + B = 9, 8 + B = 9, B = 1So, A = 8, B = 1

Page No 5.30:

Question 6:

Solve each of the following Cryptarithm:
AB7+7AB 9 8A¯

Answer:

If A + B = 8, A + B  9 is possible only if A = B = 9But from 7 + B = A, A = B = 9 is impossibleSurely, A + B = 8, A + B  9So, A + 7 = 9, Surely A = 27 + B = A, 7 + B = 2,  B = 5So, A = 2, B = 5

Page No 5.30:

Question 7:

Show that the Cryptarithm 4×AB¯=CAB¯ does not have any solution.

Answer:

0 is the only unit digit number, which gives the same 0 at the unit digit when multiplied by 4. So, the possible value of B is 0.Similarly, for A also, 0 is the only possible digit.But then A, B and C will all be 0.And if A, B and C become 0, these numbers cannot be of two-digit or three-digit.Therefore, both will become a one-digit number.Thus, there is no solution possible.



Page No 5.5:

Question 1:

Without performing actual addition and division write the quotient when the sum of 69 and 96 is divided by
(i) 11
(ii) 15

Answer:

(i) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 11, we get 15 (sum of the digits) as quotient.

(ii) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 15 (sum of the digits), we obtain 11 as quotient.

Page No 5.5:

Question 2:

Without performing actual computations, find the quotient when 94 − 49 is divided by
(i) 9
(ii) 5

Answer:

(i) We know that when ab-ba  is divided by 9, the quotient is a-b.Therefore, when (94-49) is divided by 9, the quotient is (9-4 = 5).(ii) We know that when ab-ba is divided by (a-b), the quotient is 9.Therefore, when (94-49) is divided by (9-4=5), the quotient is 9.

Page No 5.5:

Question 3:

If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.

Answer:

The sum of (985+859+598) when divided by:(i) 111Quotient = (9+8+5)=22(ii) 22, i.e, (9+8+5)Quotient = 111(iii) 37 (= 1113)Quotient = 3(9+8+5)=66

Page No 5.5:

Question 4:

Find the quotient when the difference of 985 and 958 is divided by 9.

Answer:

If abc-acb is divided by 9, the quotient is (b-c).If (985-958) is divided by 9, quotient=(8-5)=3



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