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)
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1. R could be e.g 280 * 3 = 840 or 940 * 3 = 2820 - insuff
2. R could be 899 or 900 - insuff
1+2. R should be > 934 - suff
C

Initially made silly mistake

+1 C

Made the same silly mistake on both the options.. . Picked (D)

I too picked D thinking r=870/3......

C +1

Oh nice question. Very tricky. C is the answer. The first statement could lead to multiple values for the hundreds digit. The second statement could also lead to two values for the hundreds digit.
However, together, the two statements can lead only to one value - 9 - for the hundreds digit of the 3-digit number.

Oops! Got S1 analysis incorrect, S2 correct. I picked A.

For S1, I totally ignored that the number could be 4 digit.

It is these simple, yet tricky, twists that I fear of.

Again let’s think about this for a second. First we have that ABC * 3 has a hundreds digit of 8. This ain't gonna tell us much so insuff. Now,

Second statement says that abc + 1 has hundreds digit of 9. Now the max value for tens and units digit respectively are 9 and 9. Then we only have 1 carried over to the units digit. Therefore a is either 8 or 9.

Now, let's try with both statements together. If we have 8XX * 3 = 4 as hundreds digit and we would need a carry over of 4 to reach 8 in the hundreds digit as per statement 1. This is not possible because even if b = 9 the maximum carry over we can get is 2 (29 if both units digit and tens digit are equal to 9). Therefore this option is not possible. What if hundreds digit is 9? Then we wil have that 3*9=27 so 7 for hundreds digit we would need only a carry over of 1 which is indeed possible. Hence only 9 works. C

Hope this helps
Cheers
J

But the question is saying that R is positive three-digit integer

I believe the correct answer to this question should be changed to answer choice A.The question stem says R is a 3 digit number.

Let'say R is represented by digits A,B,and C, where A is a digit between 1 and 9 inclusive and B and C are digits between 0 and 9 inclusive.

(1) R=ABC can be written as R=100*A+10*B+C. 3*R can be written as 300*A+30*B+3*C. The maximum value of 30*B+3*C is 297 when B and C are each 9.
Since it is required that 300*A+30*B+3*C > 800, if we minimize A we get, 300*A > 503. But A cannot be greater than or equal to 3, because then the number would be at least 900 and the hundreds digit of 3R will no longer be 8. So A must be 2. First statement is sufficient

(2) You can easily find this statement insufficient by taking numbers 899 and 900. The hundreds digit of R can be 8 or 9, hence insufficient.

Correct answer choice is A.

Hi VeritasPrepKarishma,

It is given that R is a 3 digit positive integer. Shouldn't the correct answer be choice A?

R is a three digit number but 3R needn't be. It could have 4 digits.
In each of the cases shown, R has 3 digits and the hundreds digit is different. So stmnt 1 is not enough.
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Yes you are right! Perhaps it is best to avoid an algebraic approach such as the one I did for a multiplication problem such as this. I will leave that algebraic approach for addition and subtraction questions.

+1 Kudos to you.

Here is an alternate method to show statement 1 is insufficient without looking at carryover. 3R is a multiple of 3, therefore its sum of digits should be a multiple of 3. In the case 3R is a 4 digit number, the thousands digit can only be 1 or 2 (the largest 3 digit number, 999, multiplied by 3 is smaller than 3000). After applying statement 1, the hundreds digit is now 8.

Let's say 3R is represented by XYZW, where X has a digit value of 1 or 2, Y has a digit value of 8, and Z and W each can be a digit between 0 and 9 inclusive.

Case 1: X=1 and Y=8

1+8+Z+W = Multiple of 3
Ex: If Z+W = 0. Then R = 600. The hundreds digit of R is 6.


Case 2: X=2 and Y=8

2+8+Z+W = Multiple of 3
Ex: If Z and W each equal 1, then 3R =2811 --> R=927. The hundreds digit of R is 9.

Statement 1 is insufficient.

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.
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