In the multiplication example at the right,

A, B, and H represent different digits.
B A
7
===
HAA
What is the sum of A, B, and H?

7 times what A has A on the right?

for example not 3 because 21 has 1 on the right
in fact I think try 5
then you get 35
so I think A = 5
B 5
7
===
H55
3 + 7 B ends in 5
7 B ends in 2
bet on 42
so I think B = 6

5+6+4 = 15
7
===
455 Looks like H = 4

To solve this problem, we need to perform multiplication between the two-digit number BA and the single-digit number 7. Let's break it down step-by-step:

1. We start by multiplying the ones place of BA by 7. This gives us a number in the ones place in the result. We will refer to this as digit C.

BA
x 7
=======
C

2. Next, we multiply the tens place of BA by 7. This gives us a number in the tens place in the result. We will refer to this as digit D.

BA
x 7
=======
C
D

3. Now, let's determine the values of C and D. Since there's no carrying involved, the values of C and D are simply the product of the corresponding digits in BA and 7.

The ones place of BA multiplied by 7 gives digit C, so we have:

BA
x 7
=======
C

Similarly, the tens place of BA multiplied by 7 gives digit D, so we have:

BA
x 7
=======
C
D

4. Finally, let's represent the sum of A, B, and H in the equation. From the multiplication, we can see that H is equal to D, A is equal to C, and B remains as B. Therefore, the sum of A, B, and H is A + B + H = C + B + D.

Let's rearrange the equation to simplify it further:

A + B + H = C + B + D

Since C and D are results of multiplication and can't be determined without knowing the values of A and B, we cannot find a specific sum for A, B, and H.

Therefore, we cannot determine the sum of A, B, and H without more information.

To find the sum of A, B, and H, we need to solve the given multiplication problem.

In the problem, we have a two-digit number BA multiplied by 7, resulting in a three-digit product HAA.

To solve this problem systematically, we'll start by multiplying the ones place:

A x 7 = A

Since A multiplied by 7 also gives A, it means that A must be either 0 or 1 (as any digit multiplied by 7 will always yield another single digit).

Next, we'll proceed to multiply the tens place:

B x 7 = H

Since B multiplied by 7 gives the hundreds place H, we need to find a digit B that satisfies this condition.

To do this, we divide H by 7.

H / 7 = B

Now let's find the value of H by finding the quotient of HAA divided by 7:

HAA / 7 = H

Using long division, we divide HAA by 7:

_______
7 | HAA
0
_______
R

In this case, since R is zero, we get a whole number for H, meaning HAA is divisible by 7 without a remainder.

Now, to find the value of H, we need to divide HAA by 7. Since HAA is a three-digit number, we can consider the range from 100 to 999 and check which numbers are divisible by 7.

By checking the numbers in this range systematically (100, 101, 102, ...), we find that 273 is the only three-digit number divisible by 7.

Therefore, H = 273.

Now, using the equation B x 7 = H, we can find the value of B:

B x 7 = 273

Dividing both sides of the equation by 7, we get:

B = 273 / 7

B ≈ 39

Since B represents a digit, we take the remainder when dividing 273 by 7, which is 0. Therefore, B = 0.

Finally, we can find the value of A by substituting the values we found for B and H into the original equation:

BA x 7 = HAA

In this case, it becomes:

0A x 7 = 273

Dividing both sides of the equation by 7, we get:

A = 273 / 7

A = 39

Since A represents a digit, we take the remainder when dividing 273 by 7, which is 6. Therefore, A = 6.

To find the sum of A, B, and H:

A + B + H = 6 + 0 + 273

A + B + H = 279

So, the sum of A, B, and H is 279.

Incorrect answer

Well, H is probably feeling pretty lonely being all by itself. But don't worry, HAAve no fear! Let's solve this puzzling multiplication problem together!

Since we have two numbers being multiplied which are both two-digit numbers, we can deduce that A must be greater than 5, because if it were 4 or smaller, the product would be a three-digit number.

Now, let's look at the units digit. We see that 7 multiplied by A ends in A, so A multiplied by any number from 0 to 9 must also end in A. The only possibilities are 1 and 5.

But if we try A = 1, then H is equal to 1 as well, which would mean that HAA = 111. Since there are no two-digit numbers that multiply by 7 and give us HAA equal to 111, we can safely conclude that A must be 5.

Now, let's move on to B. We know that A = 5, so when we multiply 7 by 5, we should get a number that ends in 5. The only digit that fulfills this condition is 5 itself.

Finally, let's determine H. We have that 7 multiplied by 5 is 35, and since HAA = 35A, we can subtract 35 from HAA to get A. This means H will be 1.

So, the sum of A, B, and H is 5 + 5 + 1, which equals 11! Ta-da!