### A problem to solve about multiples

Here's one more problem from the collection that John Morse sent me.

I chose this problem because solving it doesn't require knowing any concepts beyond multiplication and multiples.

I solved this problem kind of a "crude" way; however upon thinking my solution through, it is fairly accessible to even younger students, because it doesn't use more sophisticated concepts.

Basically I considered the problem as finding ABC (A, B, and C are digits), or possibly a longer or shorter number such as 144 x ABC ends in 144.

I systematically checked what C can be in order for the answer to end in 4.

I found only one possible digit works.

Then I systematically checked what B can be, knowing that C must equal 6 -- and found two possible digits: 2 and 7.

After that, I stumbled upon the right answer since I simply checked what is 26 x 144, 76 x 144 and 126 x 144. The last one is the multiple we're looking for - it is 18,144.

Like I said, this method IS accessible to students who have mastered multiplication algorithm.

Another solution, essentially by John Morse:

We're looking for a multiple of 144 that ends in ...144. Since 144 is a multiple of 16, ITS multiples will also be multiples of 16. Similarly, 144 is a multiple of 9, thus ITS multiples will also be multiples of 9 (divisible by 9).

Then, we do know our number will be greater than 1,000. The number we look for is (some thousands) + 144 since it must end in 144. Since 144 is a multiple of 16, those whole thousands tacked to our multiple must be also be multiples of 16. 1000 is NOT a multiple of 16, whereas 2000 (and its multiples, all having even-number thousand's place digits) ARE multiples of 16.

Hence, it remains to find the digits in front of ...144 such that they, placed together, form a multiple of 9 AND 2. The least such digit pair is 1 & 8, putting even digit 8 in the thousands's place, and thus forming 18144.

144, being a multiple of itself, naturally ends with ...144.

What is the next greater multiple of 144 ending in ...144?

I chose this problem because solving it doesn't require knowing any concepts beyond multiplication and multiples.

I solved this problem kind of a "crude" way; however upon thinking my solution through, it is fairly accessible to even younger students, because it doesn't use more sophisticated concepts.

Basically I considered the problem as finding ABC (A, B, and C are digits), or possibly a longer or shorter number such as 144 x ABC ends in 144.

144

x ABC

-----

I systematically checked what C can be in order for the answer to end in 4.

I found only one possible digit works.

Then I systematically checked what B can be, knowing that C must equal 6 -- and found two possible digits: 2 and 7.

After that, I stumbled upon the right answer since I simply checked what is 26 x 144, 76 x 144 and 126 x 144. The last one is the multiple we're looking for - it is 18,144.

Like I said, this method IS accessible to students who have mastered multiplication algorithm.

Another solution, essentially by John Morse:

This one uses the concept of a "digital root", which means essentially the remainder when dividing a number by 9. You can find it out by adding the digits of a number until you get a sum less than 10.

For example, the digital root of 28,294 is found this way: 2 + 8 + 2 + 9 + 4 = 25; 2 + 5 = 7.

When finding the digital root (or divisibility by 9), you can always "cast out nines" or any combination of numbers adding up to nine - in other words, omit them from the total sum. In the above example, it's enough to add 2 + 8 + 2 + 4 = 16; 1 + 6 = 7 to obtain the digital root.

We're looking for a multiple of 144 that ends in ...144. Since 144 is a multiple of 16, ITS multiples will also be multiples of 16. Similarly, 144 is a multiple of 9, thus ITS multiples will also be multiples of 9 (divisible by 9).

Then, we do know our number will be greater than 1,000. The number we look for is (some thousands) + 144 since it must end in 144. Since 144 is a multiple of 16, those whole thousands tacked to our multiple must be also be multiples of 16. 1000 is NOT a multiple of 16, whereas 2000 (and its multiples, all having even-number thousand's place digits) ARE multiples of 16.

Hence, it remains to find the digits in front of ...144 such that they, placed together, form a multiple of 9 AND 2. The least such digit pair is 1 & 8, putting even digit 8 in the thousands's place, and thus forming 18144.

## Comments

Dear Maria Miller,

Thank you for posting

another of my mathematics

problems! ("...144x ends

in 144 for least x>1").

I'm very pleased that the

problems I composed are

of such interest, rele-

vance, and beneficial

to learners.

Sincerely,

John W. Morse

Delmar NY USA