Areteem: MC II-B Number Theory

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MC II-B Number Theory

I need help with problem 2.29 of the MC II-B Number theory book. The questions is “How many 5 digit numbers abcde (so abcde is included in 0…9 with a not equal to 0) such that the sum of abcd and bcde is divisible by 11?”

I need help solving this problem.

This question is definitely a little tricky! As a general hint, this problem is meant to test your understanding of the divisibility rule for 11 and its proof.

Using place values, what is the sum of the two four digit numbers abcd and bcde? Then try to think about how that sum is divisible by 11. You will probably use reasoning similar to the proof of the divisibility rule for 11.

Hope this helps!

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(a+b)1000+(b+c)100+(c+d)10+d+e

(a+b)-(b+c)+(c+d)-(d+e)=a+e

It means b,c,d don't matter. so (a,e)=(2,9), (3,8),(4,7)..... so 8 different case.

8*10*10*10=8000.

answer is 9000. why?

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Overall good job! But be a little careful with your algebra. Shouldn't you get a-e instead of a+e?

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