An algebra problem by Dinesh Grover

The divisibility rule of 11 is:

If you sum every second digit and then subtract all other digits and the answer is 0 or divisible by 11.

The difference of alternative sums of digits of $45674567345$

$= ( 4 + 6 + 4 + 6 + 3 + 5 ) - (5 + 7 + 5 + 7 + 4 ) = 28 - 28 = 0$

Therefore, $\boxed {11 | 45674567345}$ .

More on divisibility rules are available on: http://www.mathsisfun.com/divisibility-rules.html

Subtract the last digit from the rest if the answer is divisible by 11 hen the number is divisible by 11.So: $4567456734-5=4567456729$ $456745672-9=456745663$ $45674566-3=45674563$ $4567456-3=4567453$ $456745-3=456742$ $45674-2=45672$ $4567-2=4565$ $456-5=451$ $45-1=44$ As 44 is $4\times11$ so $45674567345$ is divisible by 11.

45 674 567 345 make groups of 3 digits from the end add alternate groups

addition of groups at odd places 345 + 674 = 1019

addition of groups at even places 567 + 45 = 612

find the difference in two additions 1019 – 612 = 407

if the difference is divisible by 11 then given number is divisible by 11

407 is divisible by 11 so 45674567345 is divisible by 11

Add the alternate digits and subtract them . thus the alternate sum would be 28 and 28 and difference would be 0 making it divisible by 11

Digits at odd sum up to 4+.6+4+6+3+5=28
Digits at even places sum up to 5+7+5+7+4 =28 Odd sum = even sum. Divisible by 11.