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Abhishek.pitti
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If A, B and C are distinct digits such that the square of the two 11 Jun 2011, 15:43
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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69% (02:37) correct 31% (02:34) wrong based on 194 sessions

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If A, B and C are distinct digits such that the square of the two digit number AB equals the three digit number CCB, then the sum of all possible values of CCB equals

A. 555
B. 666
C. 777
D. 888
E. 999

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My doubt is that do these kind of tough problems ever pop up in the actual exam? I couldn't solve these problems even in 5 mins and i never found them in practice CATs and actual GMAT exam. Please clarify my doubt that whether I should prepare for such kind of problems and also how to tackle them in limited time frame of 2 Mins.
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KarishmaB
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If A, B and C are distinct digits such that the square of the two 18 Sep 2014, 22:33
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Responding to a pm:
Quote:

If A, B and C are distinct digits such that the square of the two digit number AB equals the three digit number CCB, then the sum of all possible values of CCB equals

a) 555
b) 666
c) 777
d) 888
e) 999

My doubt is that do these kind of tough problems ever pop up in the actual exam? I couldn't solve these problems even in 5 mins and i never found them in practice CATs and actual GMAT exam. Please clarify my doubt that whether I should prepare for such kind of problems and also how to tackle them in limited time frame of 2 Mins.
Show more

Actually, the problem isn't very tough. Certainly more tedious than I would like but 2 mins is enough to solve it.

"the square of the two digit number AB equals the three digit number CCB"
The number is AB and its square is CCB - both end with the same digit.
This happens with only 4 digits: 0, 1, 5, 6
The square of a number ending with 0 will have two zeroes at the end and will be of the format C00 i.e. CBB (not CCB as given). Hence 0 is not possible.

Units digit 1 - 11, 21, 31 etc Larger numbers' square is 4 digit so ignore them.
Square of 11 is 121 - (Not Acceptable). Square of 21 = 441 (Acceptable). Square of 31 is 961 (Not acceptable)

Unit's digit 5 - 15, 25. Larger numbers' square is 4 digit (35^2 = 1225) so ignore them.
Square of 15 is 225 (Acceptable). Square of 25 is 625 (Not Acceptable).

We already have the sum as 441 + 225 = 666

Unit's digit 6 - 16, 26
Square of 16 is 256 (Not Acceptable). Square of 26 will be much larger such that when you add it to 666, the sum will exceed 1000 (not in options).

Hence, answer must be (B)
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If A, B and C are distinct digits such that the square of the two 04 Oct 2014, 00:23
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VeritasPrepKarishma
Responding to a pm:
Quote:

If A, B and C are distinct digits such that the square of the two digit number AB equals the three digit number CCB, then the sum of all possible values of CCB equals

a) 555
b) 666
c) 777
d) 888
e) 999

My doubt is that do these kind of tough problems ever pop up in the actual exam? I couldn't solve these problems even in 5 mins and i never found them in practice CATs and actual GMAT exam. Please clarify my doubt that whether I should prepare for such kind of problems and also how to tackle them in limited time frame of 2 Mins.

Actually, the problem isn't very tough. Certainly more tedious than I would like but 2 mins is enough to solve it.

"the square of the two digit number AB equals the three digit number CCB"
The number is AB and its square is CCB - both end with the same digit.
This happens with only 4 digits: 0, 1, 5, 6
The square of a number ending with 0 will have two zeroes at the end and will be of the format C00 i.e. CBB (not CCB as given). Hence 0 is not possible.

Units digit 1 - 11, 21, 31 etc Larger numbers' square is 4 digit so ignore them.
Square of 11 is 121 - (Not Acceptable). Square of 21 = 441 (Acceptable). Square of 31 is 961 (Not acceptable)

Unit's digit 5 - 15, 25. Larger numbers' square is 4 digit (35^2 = 1225) so ignore them.
Square of 15 is 225 (Acceptable). Square of 25 is 625 (Not Acceptable).

We already have the sum as 441 + 225 = 666

Unit's digit 6 - 16, 26
Square of 16 is 256 (Not Acceptable). Square of 26 will be much larger such that when you add it to 666, the sum will exceed 1000 (not in options).

Hence, answer must be (B)
Show more

actually. it took me less than 2 mins to solve this. figured out the unit's digit be same on both LHS and RHS and then plugged in values. :)
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If A, B and C are distinct digits such that the square of the two 02 Jul 2015, 23:49
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VeritasPrepKarishma
Responding to a pm:
Quote:

If A, B and C are distinct digits such that the square of the two digit number AB equals the three digit number CCB, then the sum of all possible values of CCB equals

a) 555
b) 666
c) 777
d) 888
e) 999

My doubt is that do these kind of tough problems ever pop up in the actual exam? I couldn't solve these problems even in 5 mins and i never found them in practice CATs and actual GMAT exam. Please clarify my doubt that whether I should prepare for such kind of problems and also how to tackle them in limited time frame of 2 Mins.

Actually, the problem isn't very tough. Certainly more tedious than I would like but 2 mins is enough to solve it.

"the square of the two digit number AB equals the three digit number CCB"
The number is AB and its square is CCB - both end with the same digit.
This happens with only 4 digits: 0, 1, 5, 6
The square of a number ending with 0 will have two zeroes at the end and will be of the format C00 i.e. CBB (not CCB as given). Hence 0 is not possible.

Units digit 1 - 11, 21, 31 etc Larger numbers' square is 4 digit so ignore them.
Square of 11 is 121 - (Not Acceptable). Square of 21 = 441 (Acceptable). Square of 31 is 961 (Not acceptable)

Unit's digit 5 - 15, 25. Larger numbers' square is 4 digit (35^2 = 1225) so ignore them.
Square of 15 is 225 (Acceptable). Square of 25 is 625 (Not Acceptable).

We already have the sum as 441 + 225 = 666

Unit's digit 6 - 16, 26
Square of 16 is 256 (Not Acceptable). Square of 26 will be much larger such that when you add it to 666, the sum will exceed 1000 (not in options).

Hence, answer must be (B)
Show more

I got the answer, but can anyone explian pls. "This happens with only 4 digits: 0, 1, 5, 6".
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If A, B and C are distinct digits such that the square of the two 03 Jul 2015, 00:45
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robinpallickal
VeritasPrepKarishma
Responding to a pm:
Quote:

If A, B and C are distinct digits such that the square of the two digit number AB equals the three digit number CCB, then the sum of all possible values of CCB equals

a) 555
b) 666
c) 777
d) 888
e) 999

My doubt is that do these kind of tough problems ever pop up in the actual exam? I couldn't solve these problems even in 5 mins and i never found them in practice CATs and actual GMAT exam. Please clarify my doubt that whether I should prepare for such kind of problems and also how to tackle them in limited time frame of 2 Mins.

Actually, the problem isn't very tough. Certainly more tedious than I would like but 2 mins is enough to solve it.

"the square of the two digit number AB equals the three digit number CCB"
The number is AB and its square is CCB - both end with the same digit.
This happens with only 4 digits: 0, 1, 5, 6
The square of a number ending with 0 will have two zeroes at the end and will be of the format C00 i.e. CBB (not CCB as given). Hence 0 is not possible.

Units digit 1 - 11, 21, 31 etc Larger numbers' square is 4 digit so ignore them.
Square of 11 is 121 - (Not Acceptable). Square of 21 = 441 (Acceptable). Square of 31 is 961 (Not acceptable)

Unit's digit 5 - 15, 25. Larger numbers' square is 4 digit (35^2 = 1225) so ignore them.
Square of 15 is 225 (Acceptable). Square of 25 is 625 (Not Acceptable).

We already have the sum as 441 + 225 = 666

Unit's digit 6 - 16, 26
Square of 16 is 256 (Not Acceptable). Square of 26 will be much larger such that when you add it to 666, the sum will exceed 1000 (not in options).

Hence, answer must be (B)

I got the answer, but can anyone explian pls. "This happens with only 4 digits: 0, 1, 5, 6".
Show more

The Numbers with unit digit 0, 1, 5 or 6 are the only Numbers which result in the number with same unit digit.

e.g. 10^2 = 100
e.g. 20^2 = 400

e.g. 11^2 = 121
e.g. 21^2 = 441

e.g. 15^2 = 225
e.g. 25^2 = 625

e.g. 16^2 = 256
e.g. 26^2 = 676

i.e. If you take any Number '\(n\)' which has unit digit other than 0, 1, 5 or 6 then the Unit digit of '\(n^2\)'will NOT be same as the Unit digit of \(n\)
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If A, B and C are distinct digits such that the square of the two 04 Jan 2016, 19:17
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took me just over 3 minutes to answer the question...
B can be only 0, 1, 5, or 6.
0 - can't be otherwise, doesn't satisfy the condition that AB * AB = CCB, where C is the same digit.

When B=1
A=1 can't be as per condition, =2 =>441 (YES), 3 => 961 (NO), If more than 3 then we have 4 digit number.

when B=2:
A=1 => 144 (NO), =2 - can't be as per condition of the question, 3 OR OVER - CAN'T BE SINCE WE HAVE 4 DIGIT NUMBER.

when B=5
A=1 -> 225 (YES), 2 = 625 (NO), If more than 2, then we have 4 digit number.

when B=6.
A=1 => 256 (NO), 2 => 676 (NO), if more than 2, then we have 4 digit number.

so far, we have 441 and 225, which equal to 666.
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If A, B and C are distinct digits such that the square of the two 24 Jan 2026, 00:02
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