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X & Y
An integer grater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n.

(2) The tens digit of n is 2

Plz Explain..


(1) 24 (2 is a factor of 4), 39 (3 is a factor of 9), 48 (4 is a factor of 8) ... sufficient

(2) 22 composite, 23 not composite - not sufficient
General Discussion
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me too A.

considering A, the number can be written as

a(ak)

where a is the tens digit and ak is the unit digit, k is some constant.

= 10a + ak
= a(10+k)

this is always divisible by a and hence composite since a is not equal to 1
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A..
SI tells us that the tens digit is a factor of units digit so possible values of units digit are 4,6,8,9.....all would be even nos except no ending with digit 9, which will be 39... so all r composite....
SII.. no can be any of 21,22,23.....29..ie mix of composite and prime
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Hi Bunuel,

I got the answer by other way. However, i would like to understand your explanation on (1) where you stated that a=kb, (0<=k<=4) --> n=10b+a=10b+kb=b(10+k) --> as b>=2, n will always be composite and factor of b.

Can you please explain in detail this a=kb part? why (0<=k<=4)?

Regards,
Mitesh
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Hi Bunuel,

I got the answer by other way. However, i would like to understand your explanation on (1) where you stated that a=kb, (0<=k<=4) --> n=10b+a=10b+kb=b(10+k) --> as b>=2, n will always be composite and factor of b.

Can you please explain in detail this a=kb part? why (0<=k<=4)?

Regards,
Mitesh

(1) says that the tens digit of n, which is b, is a factor of the units digit of n, which is a. So, b is a factor of a --> \(a=kb\), for some integer k.

As for \(0\leq{k}\leq{4}\): the least value of b is 2, thus if k is 5 or more, then a becomes 10 or more, which is not possible since we know that a is a digit (\(0\leq{k}\leq{4}\)).

Hope it's clear.
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Bunuel
An integer grater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

Given: \(n>20\) --> two digit integer can be written as follows: \(n=10b+a>20\) --> \(2\leq{b}\leq{9}\), \(0\leq{a}\leq{9}\).

(1) The tens digit of n is a factor of the units digit of n. This statement implies that \(a=kb\), (\(0\leq{k}\leq{4}\)) --> \(n=10b+a=10b+kb=b(10+k)\) --> as \(b\geq{2}\), \(n\) will always be composite and factor of \(b\). Sufficient

(2) The tens digit of n is 2. This statement implies that \(b=2\), but \(n\) can be for instance composite 25 or prime 29. Not sufficient.

Answer: A.

chetan2u
A..
SI tells us that the tens digit is a factor of units digit so possible values of units digit are 4,6,8,9.....all would be even nos except no ending with digit 9, which will be 39... so all r composite....
SII.. no can be any of 21,22,23.....29..ie mix of composite and prime



Actually units digit of \(n\) can be any digit but 1, meaning that 0, 2, 3, 5, 7 too. And for 9: 99 also fits: eg . 30, 40, 50, 60, 70, 80, 90, 22, 33, 55, 77, 99.

Hi ,

What about No. 71 .
it is a prime number and
7 is factor of 1.
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Bunuel
An integer grater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

Given: \(n>20\) --> two digit integer can be written as follows: \(n=10b+a>20\) --> \(2\leq{b}\leq{9}\), \(0\leq{a}\leq{9}\).

(1) The tens digit of n is a factor of the units digit of n. This statement implies that \(a=kb\), (\(0\leq{k}\leq{4}\)) --> \(n=10b+a=10b+kb=b(10+k)\) --> as \(b\geq{2}\), \(n\) will always be composite and factor of \(b\). Sufficient

(2) The tens digit of n is 2. This statement implies that \(b=2\), but \(n\) can be for instance composite 25 or prime 29. Not sufficient.

Answer: A.

chetan2u
A..
SI tells us that the tens digit is a factor of units digit so possible values of units digit are 4,6,8,9.....all would be even nos except no ending with digit 9, which will be 39... so all r composite....
SII.. no can be any of 21,22,23.....29..ie mix of composite and prime



Actually units digit of \(n\) can be any digit but 1, meaning that 0, 2, 3, 5, 7 too. And for 9: 99 also fits: eg . 30, 40, 50, 60, 70, 80, 90, 22, 33, 55, 77, 99.

Hi ,

What about No. 71 .
it is a prime number and
7 is factor of 1.

7 is not a factor of 1, 7 is a multiple of 1.
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Please explain, why can't we consider 31( or 41) number for statement 1.

Is 3 a factor of 1?
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Please explain, why can't we consider 31( or 41) number for statement 1.

Is 3 a factor of 1?

In 31, 3 is the tens digit and 1 is the units digit. (1) says that the tens digit of n is a factor of the units digit of n but 3 is NOT a factor of 1, 1 is a factor of 3, or in another way 3 is a multiple of 1.

Hope it's clear.
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For condition one - They have mentioned that ten's digit is a factor of one's digit.
Implying that numbers are - 22 ( 2 is factor of 2 ) , 24 ( 2 is a factor of 4 ) , 26, 28, 33, 36, 39 etc. Thus these numbers are composite as they are not prime.

Condition 2 , gives us 21,22,23,24....29. Of which 23, 29 are prime . Thus insufficient.

IMO - A .
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An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n.

Lets try to find a prime number that would work in this scenario. Since the number has to be greater than 20, we can't use 11, 13, etc.

33 is a composite number, as is 77. SUFFICIENT.

(2) The tens digit of n is 2

We can get a prime number and a composite number. 23 is prime, 21 is composite. INSUFFICIENT.

Answer is A.
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(1) n = ab

n = 10a + b

If a is a factor of b then ax = b

n = a(10+x) we can see here immediately that since we can factor out a then a will serve as another factor of n. Also a cannot be equal to 1 as stated in the question stem. Thus n is composite.

SUFFICIENT.

(2) clearly insufficient (23,29)

Answer is A.
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This question makes a good argument for memorizing all the primes less than 100.

If you know all the two-digit primes in advance, then you can simply rule them out individually.

What do the numbers 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97 all have in common? The tens digit (the first digit) is not divisible by the ones digit.
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but this question talks about a single "integer" and if the statements are not boiling down to a single correct answer- isn't it still incomplete?
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ayushituteja02
but this question talks about a single "integer" and if the statements are not boiling down to a single correct answer- isn't it still incomplete?

There are two types of data sufficiency questions:

1. YES/NO DS Questions:

In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".


2. VALUE DS QUESTIONS:

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.

This question belongs to the first type. Notice that the question asks "is is n composite?" not "what is the value of n" and from the first statement we can get a definite YES answer to the question.

Hope it's clear.
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what about 91 or 71 - they both fulfill the statement 1 condition but we are getting two different result 91- composite while 71 is non-composite
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ajit257
Q13:
An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n.
(2) The tens digit of n is 2.

Not sure about the ans

The moment I hear two digit prime number, I think of numbers having 1/3/7/9 as unit's digit. A two digit number ending in 0/2/4/5/6/8 cannot be prime since they are either even or a multiple of 5 (if it is a single digit, 5 is prime)

Stmnt 1: Ten's digit is a factor of the unit's digit.
It is easy to see that there are many such composite numbers e.g. 24, 26 etc.

I will try to find if there are any such prime numbers: The only factor of 1 is 1; of 3 are 1 and 3; of 7 are 1 and 7; of 9 are 1, 3 and 9.
11, 13, 17 and 19 are not allowed since the number should be greater than 20.
33, 77, 39 and 99 are all composite. So there is no such prime number. All such number will be composite so n is composite. Hence sufficient.

Stmnt 2: The tens digit of n is 2.
21 is not prime while 23 is. Hence not sufficient.

Answer (A).
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what about 91 or 71 - they both fulfill the statement 1 condition but we are getting two different result 91- composite while 71 is non-composite
KarishmaB
ajit257
Q13:
An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n.
(2) The tens digit of n is 2.

Not sure about the ans

The moment I hear two digit prime number, I think of numbers having 1/3/7/9 as unit's digit. A two digit number ending in 0/2/4/5/6/8 cannot be prime since they are either even or a multiple of 5 (if it is a single digit, 5 is prime)

Stmnt 1: Ten's digit is a factor of the unit's digit.
It is easy to see that there are many such composite numbers e.g. 24, 26 etc.

I will try to find if there are any such prime numbers: The only factor of 1 is 1; of 3 are 1 and 3; of 7 are 1 and 7; of 9 are 1, 3 and 9.
11, 13, 17 and 19 are not allowed since the number should be greater than 20.
33, 77, 39 and 99 are all composite. So there is no such prime number. All such number will be composite so n is composite. Hence sufficient.

Stmnt 2: The tens digit of n is 2.
21 is not prime while 23 is. Hence not sufficient.

Answer (A).

(1) says: The tens digit of n is a factor of the units digit of n.

In 91, 9 is the tens digit which is not a factor of the units digit 1, it's a multiple of 1.
In 71, 7 is the tens digit which is not a factor of the units digit 1, it's a multiple of 1.
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