Number Problems

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Number Problems

by chayanika » Sat Sep 13, 2008 4:38 am
A nine digit number is such that it contains all the nine digits from 1 to 9(no repition) such that the number formed by the first 2 digit is divisible by2, the number formed by the first 3 digit is divisible by 3, ...., the number formed by the first 8 digit number is divisible by 8 and the entire number is divisible by 9. What is the number and the sum of the last 6 digits?

Ans : The number is 381654729. Sum of the last 6 digit number is 33.

Can somebody please explain how to tackle these questions.
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by niraj_a » Sat Sep 13, 2008 5:54 am
i think the fastest way to solve this question, if you realized you were taking too much time to get going, like i did, would be to work backwards from the answer choices and eliminate ones that break the divisibility rules outlined.

that's what i would do.

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by chayanika » Sat Sep 13, 2008 7:05 am
The choices for the sum of digits are : 37,35,27,33 & 36

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by Ian Stewart » Sat Sep 13, 2008 8:36 am
I don't see how you could find the exact number within two minutes, at least not without a calculator- you need to try dividing up to ten different seven-digit numbers by 7.

Anyway, you can get most of the way as follows:

-the first five-digit number is divisible by 5, so the fifth digit is 5;

-the second, fourth, sixth and eighth digits clearly must be even, since they are the last digits of even numbers. The rest of the digits must be odd, since no digit is repeated;

-the first three digits add to a multiple of 3, as do the first 6 digits, so the sum of digits four, five and six must also be a multiple of 3. Similarly, the sum of digits seven, eight and nine must also be a multiple of 3 (this lets you rule out two answer choices immediately);

-knowing that digits four and six are even, that digit five is 5, and that these digits sum to a multiple of 3, we only have two possibilities: digits four and six are either 2 and 8, or 4 and 6;

-we know that digits three and four form a two-digit number divisible by 4, and digit three is odd, which means that digit four can't be 8 or 4. So the middle three digits must be, in order, either 654 or 258;

-digits six through eight form a three-digit number which is divisible by 8. But since digit six is even, digits seven and eight alone must form a two-digit number divisible by 8. There aren't many possibilities to check; we find that if the middle digits are 258, the seventh and eighth digits can only be either 16 or 96, and if the middle digits are 654, the seventh and eighth digits can only be 72 or 32.

-now, since we know that the last three digits sum to a multiple of 3, we know the last six digits can only be:

258963
654321
654327
654723
654729

So the sum of the last six digits could only be 21, 27 or 33.

-For the number itself, using that digits three and four form a multiple of 4, we have the following possibilities:

789654321
987654321
981654327
189654327
147258963
741258963
981654723
189654723
183654729
381654729

The only fact we haven't used is that the first seven digits form a number divisible by 7. That's not easy to check quickly, but only one of the above numbers satisfies the condition- 381654729.
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