1. ## Two 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!

2. 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.
If it is divisible by 36, it is also divisible by 4, meaning the last two digits, 5B, must be divisible by 4 (B is 2 or 6).

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.

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?
$\displaystyle \frac{\frac{1}{2}+\frac{1}{3}+x}{3} = 1$

$\displaystyle \frac{1}{2}+\frac{1}{3}+x = 3$

$\displaystyle x = 2\frac{1}{6}$

Thank you!
You're welcome, though there is a button for that type of thing (hint, hint...)

3. Originally Posted by fecoupefe
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!
The number A55B must be even and the sum of its digits must be divisible by 3. Assign B=0,4,6,8 and assign A in such a manner that the sum of all 4 digits is divisible by 3 and you arrive at the number: 6552 or 2556.
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$\displaystyle \frac{\frac{1}{2}+\frac{1}{3}+x}{3}=1$

$\displaystyle \frac{1}{2}+\frac{1}{3}+x=3$

$\displaystyle x=3-\frac{1}{3}-\frac{1}{2}$

$\displaystyle x=\frac{13}{6}$

Edit: You beat me Henderson, but that's ok. I'll give you your thanks!! Hint. Hint.

4. ## Done!

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!

5. ## Re: Two homework problems

I have a question, though. Why must the sum of its digits be divisible by 3?

6. ## Re: Two homework problems

Originally Posted by TheMathQueen
I have a question, though. Why must the sum of its digits be divisible by 3?
actually, the sum of its digits must be divisible by 9 (which also makes it divisible by 3)

Why Divisibility Rules Work

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### the 4 digit number A 55 B is visible by 18 without remainder

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