+64 12ab56 is a multiple of 36. Find the sum of all possible values of a+b.
12ab56 is a multiple of 36. Find the sum of all possible values of a+b.
Since the sum of the given digits =1+2+5+6=14, therefore the sum of the 2 missing digits + 14 must be divisible by 9 and 4, since 36=2^2 * 3^2. The 2 digits that will meet that condition are:
1+3 =4, or 3 +1 =4
2+2=4
4+0, or 0+4=4
4+9, or 9+4 =13
5+8, or 8+5 =13
6+7,or 7+6 =13
So, in total you 11 intrchangeable pairs of numbers that will meet the condition specified.
12ab56 is a multiple of 36. Find the sum of all possible values of a+b.
Since the sum of the given digits =1+2+5+6=14, therefore the sum of the 2 missing digits + 14 must be divisible by 9 and 4, since 36=2^2 * 3^2. The 2 digits that will meet that condition are:
1+3 =4, or 3 +1 =4
2+2=4
4+0, or 0+4=4
4+9, or 9+4 =13
5+8, or 8+5 =13
6+7,or 7+6 =13
So, in total you 11 intrchangeable pairs of numbers that will meet the condition specified.
+9665 FYI: If a number is divisible by 9, the sum of all its digit is divisible by 9
If a number is divisible by 4, the last 2 digits is divisible by 4
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Because 56 is divisible by 4, 12ab56 must be divisible by 4 no matter what a and b is.
1 + 2 + a + b + 5 + 6 = 9x
14 + a + b = 9x
9x may be 18 or 27.
Let's first consider the case of 9x = 18.
14 + a + b = 18
a + b = 4 <---- This is easy maths.
a = 1 or b = 3
OR
a = 2 or b = 2
OR
a = 3 or b = 1
3 possibilities of a + b. Then we add them up:
3(a + b)
= 3(4) <---- because each pair of a + b in this consideration adds up to 4.
= 12.
Now consider the case of 9x = 27
14 + a + b = 27
a + b = 13 <----- This is still easy maths.
a = 4 or b = 9
OR
a = 5 or b = 8
OR
a = 6 or b = 7
OR
a = 7 or b = 6
OR
a = 8 or b = 5
OR
a = 9 or b = 4
6 possibilities of a + b. Then add them up again:
6(a + b)
= 6(13)
= 78
Now we add up the number we got in last part and we get in this part.
12 + 78 = 90. Final answer.
The actual sum of all possible combinations of the two numbers is:
5 x 4 = 20 and 6 x 13 =78
20 + 78 = 98
+118653 Thanks Max and guest :) Maybe there is some disagreement here?
So I will include my 2cents worth :)
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12ab56 is a multiple of 36. Find the sum of all possible values of a+b.
12ab56 is a multiple of 36. Find the sum of all possible values of a+b.
I assume a and b are missing digits of the 6 digit number :/
1+2+5+6+a+b = 9k
14+a+b=9k
a+b=9k-14
0<= a+b <19
if k=2 then a+b=4
SO a+b must equal 4
AND the sum of all possible values of a+b must also be 4
