Question

12 divides ab313ab (in decimal notation), where a, b are digits > 0, the smallest value of a + b is

Answer 1

answer : 7

a number is divisible by 12 means the number is also divisible by 4 and 3.

from divisibility of 4 : last two digits of any number should be divisible by 4.

so, here number is ab313ab

so, last two digits = ab must be divisible by 4. ......(1)

divisibility of 3 : sum of digits of number should be divisible by 3.

here number ab313ab ,

so, a + b + 3 + 1 + 3 + a + b = 2(a + b) + 7 must be divisible by 3. ...... ..(2)

now choose two arbitrary numbers such that these follow above two conditions.

let a = 5 and b = 2

then, then, 2(5 + 2) + 7 = 21 is divisible by 3.

and 5231352 , 52 is also divisible by 4.

so, a = 5 and b = 2

hence, the smallest value of (a + b) = 5 + 2 = 7

Answer 2

Answer:

7

Step-by-step explanation:

12 divides ab313ab (in decimal notation), where a, b are digits > 0, the smallest value of a + b is

ab313ab is divisble by 3 & 4

ab313ab is divisble by 3

=> a + b + 3 + 1 + 3 +a + b is divisible by 3

=> 7 + 2(a + b) is divisible by 3

=> a+b = 4 , 7  , 10 , 13 , 16   ( 1 is not possible as  a & b > 0 so a+b > 1)

ab313ab is divisble by 4

=> ab is divisible by 4 possible solution

16   ,  28  ,  52  , 64  , 76  ,  88 ,

1+ 6  & 5 + 2  = 7   is min

the smallest value of a + b is 7