What is the value of the two-digit positive integer n?

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

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.

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.

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.

from st2, test takers soon realize that the unit digit will be 9, and with st1, the value of n is blocked => C

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

Thanks a lot

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.

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.

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)

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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