I am having problems with these ones:

The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.

Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?

Thank you!

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- Oct 25th 2008, 06:51 AMfecoupefeTwo homework problems
I am having problems with these ones:

The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.

Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?

Thank you! - Oct 25th 2008, 08:00 AMHendersonQuote:

I am having problems with these ones:

The 4-digit number A 55 B is divisible by 36. What is the sum of A & B.

Also if it is divisible by 36, it is also divisible by 9, so A+5+5+B is divisible by 9, so A+B must be either 8 or 17.

Since B is at most 6, it is impossible for A+B to equal 17, so the sum must be 8.

Quote:

Of 3 numbers, two are 1/2 and 1/3. What should the third number be so that the average of all three is 1?

Quote:

Thank you!

- Oct 25th 2008, 08:08 AMmasters
The number

55**A**must be even and the sum of its digits must be divisible by 3. Assign**B**=0,4,6,8 and assign**B**in such a manner that the sum of all 4 digits is divisible by 3 and you arrive at the number:**A****6552 or 2556.**

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Edit: You beat me Henderson, but that's ok. I'll give you your thanks!! Hint. Hint. - Oct 25th 2008, 08:17 AMQuantumFractalsDone!
1) For the first question I cant think of a particular method apart from brut force. U know its divisible by 36, therefore it must be in the 36 times table. You know it is a number that contains four elements. So just list all the numbers that fit this category. i.e. 1008-9976.

Ul find that the numbers that come of the form A55B, are 2556 & 6552. Therefore in both cases A + B =__8!__

2) There is a much more methodical approach to this problem. For the average to be equal to one, you need the sum of all 3 numbers to = 3. so far u have 1/2 and 1/3 as the numbers. you need to put them under a common denominator, so 1/2 = 3/6 and 1/3=2/6. So remember you need the total to be 3 = 18/6. So your remaining number is simply 18/6 - 5/6 =__13/6!__ - Sep 8th 2016, 09:10 PMTheMathQueenRe: Two homework problems
I have a question, though. Why must the sum of its digits be divisible by 3?

- Sep 9th 2016, 06:24 AMskeeterRe: Two homework problems
actually, the sum of its digits must be divisible by 9 (which also makes it divisible by 3)

Why Divisibility Rules Work