Problem Help (AB AB + AB =19B) Need to show work, not just answer
- Thread starter jldaven
- Start date
D
Deleted member 4993
Guest
What are your thoughts?
Please share your work with us ...even if you know it is wrong.
If you are stuck at the beginning tell us and we'll start with the definitions.
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Revised Question
In the addition problem below, the letters AB represent a two-digit number. If you know that the letter B is not a zero (0), can you tell me which numbers represent A and B?
AB
AB
+ AB
=19B
Need to show work.
Thank you!
I am a parent and I've figured out the answer but I think there is an easier way to do it. I'm trying to explain it to my 4th grade son. Here is what I've come up with
Answer
65
65
65
= 195
How I figured it out:
199 / 3 =66.33 (no)
198 /3 = 66 (no - A & B can't be the same)
197 / 3 = 65.67 (no)
196 / 3 =65.33 (no)
195 / 3 = 65 (yes - A=6 & B = 5)
There must be an easier way???
Thank you!
In the addition problem below, the letters AB represent a two-digit number. If you know that the letter B is not a zero (0), can you tell me which numbers represent A and B?
AB
AB
+ AB
=19B
Need to show work.
Thank you!
I am a parent and I've figured out the answer but I think there is an easier way to do it. I'm trying to explain it to my 4th grade son. Here is what I've come up with
Answer
65
65
65
= 195
How I figured it out:
199 / 3 =66.33 (no)
198 /3 = 66 (no - A & B can't be the same)
197 / 3 = 65.67 (no)
196 / 3 =65.33 (no)
195 / 3 = 65 (yes - A=6 & B = 5)
There must be an easier way???
Thank you!
D
Deleted member 4993
Guest
My way:
B + B + B = 10 * x + B (where x is another integer) → B = 5*x → B = 5 (since B is a single digit integer)
19B = 190 + B = 195
3*(10*A + 5) = 195
30 A = 180 → A = 6
D
Deleted member 4993
Guest
For a fourth grader - your method combined with Denis's method would be more understandable.
- Joined
- Feb 4, 2004
- Messages
- 16,582
I will guess that "19B" is supposed to be a three-digit number, so the exercise is as follows:
Code:
Find the values of "A" and "B" in the following addition:
A B
A B
+ A B
-----
1 9 BYou know that B + B + B = B or else 10 + B (if there's a carry of "1") or 20 + B (if there's a carry of "2"). It can't be B + B + B = B, because this would require that B = 0, which is not allowed. So either B + B + B (= 3B) = 10 + B or else B + B + B _= 3B) = 20 + B.
Try 10 + B: You know that 3*3 = 9, which is too small. You know that 3*4 = 12, 3*5 = 15, and 3*6 = 18. Of these, only 3*5 = 10 + 5 will work.
Try 20 + B: You know that 3*6 = 18, which is too small. You know that 3*7 = 21, 3*8 = 24, and 3*8 = 27. None of these is of the form 20 + B, so this option cannot work.
So now you know that it must be that B = 5. Then you have:
Code:
Using B = 5, so 3*B = 15:
1
A 5
A 5
+ A 5
-----
1 9 5And so forth.
