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KarishmaB
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clearly statement 1 leads to many options and same to statement 2

now taking both the statements together.

it should be a no. which is common multiple of both 5 and 9 and also is 2 digit no. which has 10th place digit as reminder...
we have 9x5=45 so for reminder to be 4 the no. should be 49, which gives us reminder as 4. no other no. satisfies all the criteria mentioned in question.

clearly option C is answer
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alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

we really dont need any calculation in this question. This is quite conceptual. we know remainder of any number when divided by 5 can only be 1,2,3 or 4.

Its given remainder equals to tens digit. we'll take the four remainders one by one.

1 >> we know tens digit should be 1 so number could only be 11 OR 16.
2 >> we know tens digit should be 2 so number could only be 22 OR 27
3 >> we know tens digit should be 3 so number could only be 33 OR 38.
4 >> we know tens digit should be 4 so number could only be 44 OR 49.

we cant get an answer from st. 1.

we can do same reasoning for st. 2
1 >> 10,19
2 >> 20,29
3 >> 30,39
4 >> 40,49 .... so on .. also no point going forward. we found a match from st. 1(and in st. 1 we wrote all the possible outcomes, so possibility of another such no. is zero) .. the no. is 49.

hence C.
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alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

Let's take each statement at a time.

(1) When n is divided by 5, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 5 - say 15. The remainder should be 1 so say n = 16.
Think of another number which is divisible by 5 - say 25. The remainder should be 2 so say n = 27
There will be more such numbers so we can see that this is certainly not sufficient.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 9 - say 18. The remainder should be 1 so say n = 19.
Think of another number which is divisible by 9 - say 27. The remainder should be 2 so say n = 29
There will be more such numbers so we can see that this is certainly not sufficient.

What do we do when we consider both statements together?
We need to think of a number divisible by both 5 and 9, say 45 (their LCM). The remainder should be 4 so add 4 to 45 to get n = 49
Think of another number divisible by both which will be the next multiple of 45 i.e. 90. The remainder should be 9 but when we divide a number by 5, the remainder cannot be greater than 4. So n cannot be 99.
Hence, there is only one such two digit number i.e. n = 49.

Answer (C)

Hi Karishma,

I'm a bit stucked with both statements together. How do you know that the remainder has to be 4 and not 1,2 or 3?

Could you please elaborate on this?

Many thanks!
Cheers
J
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alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

Let's take each statement at a time.

(1) When n is divided by 5, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 5 - say 15. The remainder should be 1 so say n = 16.
Think of another number which is divisible by 5 - say 25. The remainder should be 2 so say n = 27
There will be more such numbers so we can see that this is certainly not sufficient.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 9 - say 18. The remainder should be 1 so say n = 19.
Think of another number which is divisible by 9 - say 27. The remainder should be 2 so say n = 29
There will be more such numbers so we can see that this is certainly not sufficient.

What do we do when we consider both statements together?
We need to think of a number divisible by both 5 and 9, say 45 (their LCM). The remainder should be 4 so add 4 to 45 to get n = 49
Think of another number divisible by both which will be the next multiple of 45 i.e. 90. The remainder should be 9 but when we divide a number by 5, the remainder cannot be greater than 4. So n cannot be 99.
Hence, there is only one such two digit number i.e. n = 49.

Answer (C)

Hi Karishma,

I'm a bit stucked with both statements together. How do you know that the remainder has to be 4 and not 1,2 or 3?

Could you please elaborate on this?

Many thanks!
Cheers
J

The tens digit of 45 is 4.
45 is the first positive two digit number which is divisible by both 5 and 9.
So when n is divided by 5 or 9, the remainder should be 4 so n should be 49. The remainder will be 4 which is the tens digit of 49.
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PROMPT ANALYSIS
N is a 2 digit number

SUPERSET
The value of n could be in the range 10 to 99

TRANSLATION
In order to find the value of n, we need:
1# Exact value of n
2# equations to find the value of n.

STATEMENT ANALYSIS
St 1: we can say that n could be 11, 22, 33, 44. INSUFFICIENT. Hence option a and d eliminated.
St 2: we can say that n could be 19, 29, 39, 49, 59, 69, 79, 89. INSUFFICIENT. Hence option b eliminated.

St 1 & St 2: there is no number in common which can follow both the statements. INSUFFICIENT.

Option E.
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from st2, test takers soon realize that the unit digit will be 9, and with st1, the value of n is blocked => C
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surya167
alexpavlos
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

Take the LCM of 9 and 5 which is 45. Since remainder is tens digit, add 4 to 45 = 49

Now, 49/5 = 9 + remainder ->4
49/9 = 5 + remainder ->4

fantastic!!! Its very simple. I wish I had thought like that..

Thanks a lot
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VeritasKarishma
alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

Let's take each statement at a time.

(1) When n is divided by 5, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 5 - say 15. The remainder should be 1 so say n = 16.
Think of another number which is divisible by 5 - say 25. The remainder should be 2 so say n = 27
There will be more such numbers so we can see that this is certainly not sufficient.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 9 - say 18. The remainder should be 1 so say n = 19.
Think of another number which is divisible by 9 - say 27. The remainder should be 2 so say n = 29
There will be more such numbers so we can see that this is certainly not sufficient.

What do we do when we consider both statements together?
We need to think of a number divisible by both 5 and 9, say 45 (their LCM). The remainder should be 4 so add 4 to 45 to get n = 49
Think of another number divisible by both which will be the next multiple of 45 i.e. 90. The remainder should be 9 but when we divide a number by 5, the remainder cannot be greater than 4. So n cannot be 99.
Hence, there is only one such two digit number i.e. n = 49.

Answer (C)

Quote:
can you elaborate as to how 99 cannot be the number. I did not understand that part

When you divide a number by any divisor say D, the remainder MUST be less than D.
Say if you divide a number by 5, can the remainder be 7? No. Then 5 will go another time in it and the remainder will be 2. So remainders can be 0/1/2/3/4 only.

Since N when divided by 5 gives remainder which is tens digit of N, the tens digit of N must be one of 0/1/2/3/4 only. So it cannot be 9 and hence 99 cannot be N.
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VeritasKarishma
alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.



Any help on this one would be much appreciated! Thanks

Let's take each statement at a time.

(1) When n is divided by 5, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 5 - say 15. The remainder should be 1 so say n = 16.
Think of another number which is divisible by 5 - say 25. The remainder should be 2 so say n = 27
There will be more such numbers so we can see that this is certainly not sufficient.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.
Think of a two digit number which is divisible by 9 - say 18. The remainder should be 1 so say n = 19.
Think of another number which is divisible by 9 - say 27. The remainder should be 2 so say n = 29
There will be more such numbers so we can see that this is certainly not sufficient.

What do we do when we consider both statements together?
We need to think of a number divisible by both 5 and 9, say 45 (their LCM). The remainder should be 4 so add 4 to 45 to get n = 49
Think of another number divisible by both which will be the next multiple of 45 i.e. 90. The remainder should be 9 but when we divide a number by 5, the remainder cannot be greater than 4. So n cannot be 99.
Hence, there is only one such two digit number i.e. n = 49.

Answer (C)

VeritasKarishma
I didn't understand why remainder cannot be greater than 4. If our new divisor is 45, then shouldn't remainder be < 45?
I mean, we aren't dividing 99 by 5, we are dividing 99 by 45 (LCM of 9 and 5), right?
In that case, we still get a remainder of 9, which is the tens digit of n = 99.

I'm sure there's a gap in my understanding, but unable to analyse what it is exactly.
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Conceptual solution

(1) n=5q+d; 10d+u=5q+d; u=5q-9d; d=5q/9-u/9. We know d=(1,2,3 or4). Not sufficient.

(2) n=9p+d; 10d+u=9p+d; u=9(p-d). --> u=0 or 9. d=p-u/9. Not sufficient.

(1)+(2) test u=0: by (1), then d=5*(Q/9). Q/9 must be integer. d cannot be multiple of 5 (can't be 0, can't be 5). u can't be 0 -> u=9.
by (2), d=p-1; d=(5/9)*q-1 --> p=5/9q; d=5q/9-1. d must be bigger than 0, lower than 5. -> q/9 must be equal to 1 --> 5q/9=5. d=5-1=4.

49. (C)
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alex1233
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.
(2) When n is divided by 9, the remainder is equal to the tens digit of n.

Algebraic approach:

Any two-digit integer can be represented as 10T+U, where T = the tens digit and U = the units digit.
Thus:
n = 10T+U

Statement 1: When n is divided by 5, the remainder is equal to the tens digit of n.
In other words, n is equal to a multiple of 5 plus a remainder of T:
n = 5a+T

Since n = 10T+U and n = 5a+T, we get:
10T+U = 5a+T
U = 5a-9T

Case 1: T=1, implying that U = 5a-9
If a=2, then U=1 --> n = TU = 11
If a=3, then U=6 --> n = TU = 16
Since n can be different values, INSUFFICIENT.

Statement 2: When n is divided by 9, the remainder is equal to the tens digit of n.
In other words, n is equal to a multiple of 9 plus a remainder of T:
n = 9b+T

Since n = 10T+U and n = 9b+T, we get:
10T+U = 9b+T
U = 9b-9T
U = 9(b-T)

Since digit U must be a nonnegative integer no greater than 9, the equation above implies that b-T=0 and U=0 or that b-T=1 and U=9.

Case 1: U=0 and b-T=0, implying that T=b
If T=b=1, then n = TU = 10
If T=b=2, then n = TU = 20
Since n can be different values, INSUFFICIENT.

Statements combined:
Case 1: U=0
Since U=0 and U = 5a-9T, we get:
0 = 5a-9T
9T = 5a
\(T = \frac{5a}{9}\)
Here, the smallest option for T occurs when a=9, with the result that T=5.
Since remainder T must be less than the divisor of 5 in Statement 1, Case 1 is not viable.

Case 2: U=9
Since U=9 and U = 5a-9T, we get:
9 = 5a-9T
9T = 5a - 9
\(T = \frac{5a}{9} - 1\)

Since remainder T must be less than the divisor of 5 in Statement 1, the equation above implies that a=9 and \(T = \frac{5*9}{9} - 1 = 4\)
Since T=4 and U=9, n = 49
SUFFICIENT.

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