CAT 2025 Lesson : Divisibility - 7, 13 and Composite

Is $3,594,248$ divisible by $7 \space \text{or} \space 13$? If not, find the remainder.

Solution

Let $x = 3,594,248 = (3 \times 10^6) + (594 \times 10^3) + (248 \times 10^6)$

When sum of alternating sets of $3$ digits starting from the right is subtracted from the sum of the rest, we get $(248 + 3) - 594 = -343$

$\text{Rem} \left(\dfrac{x}{7} \right) = \text{Rem} \left(\dfrac{-343}{7} \right) = 0$

$\text{Rem} \left(\dfrac{x}{13} \right) = \text{Rem} \left(\dfrac{-343}{13} \right) = \text{Rem} \left(\dfrac{-5}{13} \right) = -5 + 13 = 8$

∴ $x$ is divisible by $\bm{7}$ and leaves a remainder of $\bm{8}$ when divided by $13$.

If $a = 92451348$, which of the following is the largest number that perfectly divides $a$?

(1) $132$            (2) $198$            (3) $396$            (4) $792$           

Solution

$a = 92451348$

From the options, taking the largest number $792$,
$792 = 2^3 \times 3^2 \times 11 = 8 \times 9 \times 11$

The last $3$ digits of $a$, $348$ is not divisible by $8$.
∴ $a$ is not divisible by $\bm{792}$.

$396 = 2^2 \times 3^2 \times 11 = 4 \times 9 \times 11$

Last $2$ digits of $a$, $48$ is divisible by $4$.
∴ $a$ is divisible by $\bm{4}$.

Sum of digits $= 9 + 2 + 4 + 5 + 1 + 3 + 4 + 8 = 36$
Sum of digits of $36 = 3 + 6 = 9$
∴ $a$ is divisible by $\bm{9}$.

Difference of alternate digits $= (2 + 5 + 3 + 8) - (9 + 4 + 1 + 4) = 0$
∴ $a$ is divisible by $\bm{11}$.
∴ $a$ is divisible by $\bm{396}$.

Answer: ($3$) $396$