VeritasKarishma wrote:
kannn wrote:
Given that R is positive three-digit integer, what is the hundreds digit of R?
1. The hundreds digit of 3R is 8
2. (R+1) results in a number with the hundreds digit of 9.
Source : 800score.com
Responding to a pm:
1. The hundreds digit of 3R is 8
When you multiply a number by 3, the hundreds digit will be obtained by multiplying the hundreds digit of the original number by 3 and adding any carryover we might have. 3 * 9 (the largest digit) is 27 so the maximum carryover we can get from tens digit multiplication is 2 (even if the units digit of the original number is also 9, we will get 2 as carryover from units to tens and still 2 as carryover from tens to hundreds).
No carry over: 6 times 3 is 18.
611 ---> 611 * 3 = 1833
One carry over: We need to obtain 7 in the hundreds place. 9 times 3 is 27.
961 ---> 961 * 3 = 2883
Two carry over: We need to obtain 6 in hundreds place. 2 times 3 is 6.
291 ---> 291 * 3 = 873
The hundreds digit can be 2, 6 or 9.
2. (R+1) results in a number with the hundreds digit of 9.
R can be an number from 899 to 998.
The hundreds digit can be 8 or 9.
Using both statements, we see that the only common hundreds digit we have is 9. Hence the hundreds digit must be 9.
Answer (C)
Quote:
For this question, what do you mean 1 carry over and 2 carry over, and how you get those numbers?
Say R = XYZ
What is 3R?
3R = 3 * XYZ
Say XYZ = 456
How will you multiply it by 3?
3 * 456 =>
3 times 6 is 18 so we write 8 in units digit and 1 carry over. We get ......8
3 times 5 is 15. Add 1 from carry over to get 16. Now write 6 in tens digit and 1 carry over. We get .......68
3 times 4 is 12. Add 1 from carry over to get 13. Now write 13. We get 1368
So how did we get the 3 in the hundreds place of 1368? We multiplied 3 with the hundreds digit (X = 4) to get 12. To that we added 1 (from carry over) and we got 13. So 3 came in our hundreds place.
This is exactly what we are doing. From previous calculations, the carry over can be 0/1/2 (explained above).
So we take each case. If carry over were 0, then R could have been something like 611 or 622 or 613 etc. Why? Because to get 8 in hundreds place, 3 needs to be multiples by 6. So 6 should be X. Y and Z should be numbers which give no carry over complication so 1/2/3 work.
We did the same thing for 1 carry over and 2 carry over too.
_________________
Karishma
Veritas Prep GMAT Instructor
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