The least value to be given to * so that the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a mult

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.Let the missing digit be $x$.The number is $5x3457$.Sum of digits at odd places (from right) $= 7 + 4 + x = 11 + x$.Sum of digits at even places (from right) $= 5 + 3 + 5 = 13$.For the number to be divisible by $11$,the difference must be $0$ or a multiple of $11$.$(11 + x) - 13 = 0$ or $(11 + x) - 13 = 11$.Case $1$: $x - 2 = 0 \implies x = 2$.Case $2$: $x - 2 = 11 \implies x = 13$ (not possible as $x$ must be a single digit).Therefore,the least value is $2$.

The least value to be given to $$ so that the number $53457$ is divisible by $11$ is

Similar Questions

$1599 \times 199 \div 49 - 1398 + 3877 = ?$

Thrice the square of a natural number decreased by $4$ times the number is equal to $50$ more than the number. The number is

$\frac{601}{49} \times \frac{399}{81} \div \frac{29}{201} = ?$

The reciprocal of the sum of the reciprocals of $\frac{6}{5}$ and $\frac{3}{7}$ is:

$(\sqrt{2209}-12) \times 5 = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The least value to be given to $*$ so that the number $5*3457$ is divisible by $11$ is