Question: Sum of 4 Unknown Digits

Comment on Sum of 4 Unknown Digits

How about we use A=9, C=-1
greenlight-admin's picture

Nice idea, but there are no negative digits.
There are 10 digits altogether: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

To hard to do in a short time. Practice, practice and practice. Thanks.
greenlight-admin's picture

Agreed! It's a tricky question.

why cant c=2?
greenlight-admin's picture

Once we know that B = 1 and D = 0, it MUST be the case that C = 9

If it were the case that C = 2, then we must be able to assign different values for A, B, C and D so that the original sum works. Can you do that with C = 2?

Cheers,
Brent

Hey I didn't get why and how you assigned B=1 so in short I'm not able to grasp the whole idea of this type of question. Please explain.
greenlight-admin's picture

The question tells us that a 2-digit number PLUS another 2-digit number yields a 3-digit SUM.

Let's examine some examples of when this occurs:
50 + 80 = 130
29 + 72 = 101
99 + 99 = 198
63 + 71 = 134
18 + 85 = 103

Notice that, in all of these cases, the hundreds digit is 1.
In fact, in this situation, the 3-digit number will ALWAYS have hundreds digit 1.

For this reason, we know that B = 1

Does that help?

Cheers,
Brent

When you say, "In fact, in this situation, the 3-digit number will ALWAYS have units digit 1." did you mean 3-digit number will always have hundreds digit 1?
greenlight-admin's picture

Oops my bad. I changed my response above to have "hundreds" (not unit)

Thanks for the heads up!

My logic to remember:

The sum of any two 2-digit numbers can never exceed 198, so, the first digit of the sum must be 1.
greenlight-admin's picture

That's perfectly logical reasoning!

Hello brent,
Can you please let me know, where can I find more similar practice questions?

Hi dear Brent,
Could you please give us more than three questions of this type, at least 10 practice questions?
Thanks a ton

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