aab+bab+bbc+acc=2860 what does a ,b ,and c equal?

Question

im just doing my homework and i nead help

Answer 1

When you write "aab", I am assuming you mean a three digit number with a in the hundreds and tens place, and b in the ones place. In that case, here is how you solve the problem:

First, if you only consider the leftmost digits, you see that a + b + b + a cannot equal more than 28. Also, if you notice that 99 + 99 + 99 + 99 = 396, you can conclude that ab + ab + bc + cc cannot equal more than 396. Thus, the right two digits can only add up to 3 in the hundreds place of the sum. In other words, the leftmost digits a + b + b + a cannot add up to more than 28, but cannot add up to less than 25, either. Also, note that a + b + b +a = 2a + 2b, so it must be even. Thus we can conclude that a + b + b + a = 26 or 28, and therefore a + b = 13 or 14.

Secondly, we note that, if a + b = 13, then a00 + b00 + b00 + a00 =2600, so the rightmost digits ab + ab + bc + cc would have to add up to 260. If a + b = 14, then a00 + b00 + b00 + a00 = 2800, so the rightmost digits ab + ab + bc + cc would have to equal 60. But this would mean that a0 + a0 + b0 + c0 would have to be less than or equal to 60, which is impossible because a0 + b0 = 140. Thus a + b cannot equal 14, so we conclude that a + b = 13.

Because a + b = 13, we now know that ab + ab + bc + cc = 260. This means that a0 + a0 + b0 + c0 is not more than 260, or that 2a + b + c is less than or equal to 26. Since a + b = 13, we see that 2a + b + c = a + 13 + c is less than or equal to 26. Thus a + c is less than or equal to 13, and since a + b = 13, c is less than or equal to b.

Now look at the ones digit. Since 9 + 9 + 9 + 9 = 36, we can conclude that b + b + c + c must equal 10, 20, or 30. Thus, b + c must equal 5, 10, or 15.

Case b + c = 5: Then b + b + c + c = 10. Since ab + ab + bc + cc = 260, we see that a0 + a0 + b0 + c0 = 250. a + b = 13, so a + c must equal 12. Now we have a system of linear equations:
a+b=13
a+c=12
b+c=5
Since b = 5 - c, a + b = a + 5 - c = 13, so a - c is 8. Since a + c is 12, a must equal 10 and c 2. But this is impossible, since a must be a digit between 0 and 9. Thus, b + c cannot = 5, and must equal 10 or 15 instead.

Case b + c = 10: Then b + b + c + c = 20. Since ab + ab + bc + cc = 260, we see that a0 + a0 + b0 + c0 = 240. a + b = 13, so a + c must equal 11. Now we have a system of linear equations:
a+b=13
a+c=11
b+c=10
Since b = 10 - c, a + b = a + 10 - c = 13, so a - c is 3. Since a + c is 11, a must equal 7 and c 4. Then b = 6. This is a potential solution.

Case b + c = 15: Then b + b + c + c = 30. Since ab + ab + bc + cc = 260, we see that a0 + a0 + b0 + c0 = 230. a + b = 13, so a + c must equal 10. Now we have a system of linear equations:
a+b=13
a+c=10
b+c=15
Since b = 15-c, a + b = a + 15 - c = 13, so a - c is -2. Since a + c is 10, a must equal 4 and c 6. Then b must equal 9.

Thus we have two possible solutions: a = 7, b = 6, c = 4; and a = 4, b = 9, and c = 6. Let's check them:

776 + 676 + 664 + 744 = 2860, check.
449 + 949 + 996 + 466 = 2860, check.

Thus, you have your two solutions.

Answer 2

i have no idea expet for this
a2b+b2a+b2c+a2c