Let N = a2607b be a six digit number divisible by 75, then 'a' can not be equal to? (A) 1 (B) 2 (C) 4 (D) 7

Nov 18, 2016

Option (B)2

Explanation:

This can be solved by applying divisibility rule as follows .

Given number $N = a 2607 b$

This is to be divisible by 75 that means by 3 as well as 25

We know by divisibility rule that for any number divisible by 25 the number formed by the last two digits of the given number should be divisible by 25. In our case the number formed by last two digit should be 75 (i.e$. b = 5$)

Now we get the number$\text{ } a 26075$ which is divisible by 25.

Now we are to check the divisibility of $\text{ } a 26075$ by 3.

To fulfill this condition the divisibility rule says that sum of digits of $\text{ } a 26075$ should be divisible by 3

This means
"a+2+6+0+7+5=a+20 should be divisible by 3.

Among the given values of a, ($1 , 2.4 .7$) only '2' can not satisfy the condition.

As for a= 1 ,$\text{ "a+20=21->color(green)"divisible by 3}$

$\textcolor{red}{\text{for a= 2, " a+20=22->color(red)"not divisible by 3}}$

for a= 4 ,$\text{ "a+20=24->color(green)"divisible by 3}$

for a= 7,$\text{ "a+20=27->color(green)"divisible by 3}$