12 divides, ab313ab (in decimal notation, where a,b are digits>0, the smallest value of a+b is
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ab313ab is divisible by 12 ,means ab313ab is divisible by 3 and 4 .
c
condition 1 :- we know, any number is divisible by 3 when sum of all digits of number is divisible by 3.
so, a + b + 3 + 1 + 3 + a + b = 7 + 2(a + b)
therefore, 7 + 2(a +b) is integral multiple of 3.
e.g., 7 + 2(a+ b) = 3n
if n = 3 , (a + b) = 1
if n = 5 , (a + b) = 4
if n = 7, (a + b) = 7
hence, (a + b) = 1 + 3P , where P = {0, 1, 2, 3...}
condition 2:- now, one more condition is ab313ab is divisible by 4 , it is possible only when last two digit ab is divisible by 4
a , b > 0 , so we have to choose a and b in such a way that it a , b > 0 , (a + b) = 1+ 3P, where P is whole number and ab is divisible by 4.
we can choose a = 1 & b = 6 , a = 5 & b = 2, etc..
so, smallest value of (a + b) = 1 + 6 or (5+ 2) =7
hence, answer should be 7
c
condition 1 :- we know, any number is divisible by 3 when sum of all digits of number is divisible by 3.
so, a + b + 3 + 1 + 3 + a + b = 7 + 2(a + b)
therefore, 7 + 2(a +b) is integral multiple of 3.
e.g., 7 + 2(a+ b) = 3n
if n = 3 , (a + b) = 1
if n = 5 , (a + b) = 4
if n = 7, (a + b) = 7
hence, (a + b) = 1 + 3P , where P = {0, 1, 2, 3...}
condition 2:- now, one more condition is ab313ab is divisible by 4 , it is possible only when last two digit ab is divisible by 4
a , b > 0 , so we have to choose a and b in such a way that it a , b > 0 , (a + b) = 1+ 3P, where P is whole number and ab is divisible by 4.
we can choose a = 1 & b = 6 , a = 5 & b = 2, etc..
so, smallest value of (a + b) = 1 + 6 or (5+ 2) =7
hence, answer should be 7
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