Prove that 4 digit palindrome is always divisible by 11
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Step-by-step explanation:
Let a be the one’s digit and b be the tenth’s digit. Then hundred’s and thousand’s digit will be b and a respectively
Then the number will be
1000a+100b+10b+a
=1001a +110b
=11(91a+10b)
So the number is divisible by 11
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