Page 15 - Arithmetic

231: The sum of numbers at odd places = 2 + 1 = 3. The sum of the numbers at even places = 3. Difference = 3 – 3 =0. So, the number is divisible by 11. So, this is not a prime number.
253: The sum of numbers at odd places = 2 + 3 = 5. The sum of the numbers at even places = 5. Difference = 5 – 5 =0. So, the number is divisible by 11. So, this is not a prime number.
9. 45657t4
A number is divisible by 4 when the last two digits are divisible by 4.
So, the least number could be 0 as the last two digits would become 04 and its divisible by 4.
10. 238a9b, for a number to be divisible by 5, the last digit should either be 0 or 5.
Case I: b = 0, the number is 238a90
Case II: b = 5, the number is 238a95
Now, for a number to be divisible by 3, the sum of the digits should be divisible by 3.
Case I: 2 + 3 + 8 + a + 9 + 0 = 22 + a, the least value of ‘a’ should be 2 for the sum to be divisible by 3.
Case II: 2 + 3 + 8 + a + 9 + 5 = 27 + a, the least value of ‘a’ should be 0 for the sum to be divisible by 3.
So, the answer can be 2, 0 or 0, 5
As per the answer option, the value of ‘a’ and ‘b’ should be 0 and 5, respectively.
11. 57740 × 98P2; for a number to be divisible by 12, the number should be divisible by 4 and 3. 57740 is divisible by 4 because the last two digits ‘40’ are divisible by 4.
Now, 98P2 should be divisible by 3.
For a number to be divisible by 3, the sum of the digits should be divisible by 3.
98P2 = 9 + 8 + P + 2 = 19 + P The least value of P should be 2.
12. For a number to be divisible by 15, it should be divisible by both 5 and 3.
And, as per the divisibility rule of 5, the number should either end in 0 or 5.
652819 ends in 9. So, subtract 4 from the number such that it becomes divisible by 5. The new number is 652815.
13. Divide 221 by 17; the remainder is 4.
14. Dividend = ( Divisor × Quotient) + Remainder Divisor = 6Q = 3R
R = 18
Divisor=3 × 18 = 54
6Q = 54
Q=9
Dividend = (54 × 9) + 18 = 504
Page 14 of 55
Arithmetic Concept