Test: Remainders- 3 - GRE MCQ

Step 1: Question statement and Inferences

We have to find the value of T

Step 2: Finding required values

Given: positive integer S is divided by a positive integer T, a remainder of 12 is obtained.

• S/T = 21.4
• Since the remainder is 12, when S is divided by T, therefore,
  • S can be written as S = T x k + 12
  • or, S/T = k + 12/T...................(i)
• And we are given that 
  • S/T = 21.4 = 21 + 0.4................(ii)
  • Comparing equation (i) and (ii), we can write -
  • 12/T = 0.4 

Step 3: Calculating the final answer

• 0.4 * T = 12
• T = 30

Answer: Option (C)

Let E = n (2n-2) (n+1)² = 2n(n-1)(n+1)²

Given: n is odd

Now n may be expressed as:

n = 2k + 1, where k is an integer

• E = 2(2k+1)(2k+1-1)(2k+1+1)²
• E = 2(2k+1)(2k)(2k+2)²
• E = 4k(2k+1)[2(k+1)]²
• E = 16k(2k+1)(k+1)²
• Thus, the given expression is divisible by 16.

Now we need to find the remainder when the expression is divided by 16 × 2 = 32.

E = 16k(2k+1)(k+1)²

We should consider two possibilities with regards to k being odd or even.

1)      k is even

Let k = 2m , where m is an integer

• E = 16(2m)[2(2m)+1](2m+1)²
• E = 32m(4m+1)(2m+1)²

Therefore E is divisible by 32.

2)      k is odd

Let k = 2m + 1, where m is an integer

• E = 16(2m + 1)[2(2m + 1)+1](2m + 1+1)²
• E = 16(2m + 1)(4m + 2+1)(2m + 2)²
• E = 16(2m+1)(4m + 3)[2(m+1)]²
• E = 64(2m+1)(4m + 3)(m+1)²

Thus, irrespective of whether k is even or odd, the given expression E will be divisible by 32.

Solution: When n(2n-2)(n+1)² is divided by 32, the remainder is zero.

Answer: Option (A)

As per the Binomial Theorem, an expression of the form (a-b)ⁿ opens up as follows:

The interesting thing to note here is that the sum of powers of a and b in each term of this expression will be equal to n. For example, if in a term, a is raised to the power n-1, then b will be raised to the power 1 in that term, because (n-1) +1 = n

Similarly, if b is raised to the power n in a term, then a will be raised to the power zero in that term.

Please don’t feel intimidated by the Binomial Theorem. Do you need to know it to crack the GMAT? The answer is No, Not really. We have discussed this expression here just to help you better visualize what’s going on here.

The point that we really want to drive home is that when the expression (a-b)ⁿ is opened up, every term except the last term has a in it.  In fact you can try it out with n = 2 or n = 3 and see it for yourself. 

This means, an expression of the form (a-b)ⁿ can be simplified as ka + (-b)ⁿ , where k is a complicated expression that is outside the scope of the GMAT.

Some students are able to figure out the above simplification intuitively, without any knowledge of the Binomial Theorem, and that is absolutely fine.

Solving the given question using Binomial Theorem

To find the remainder, we should be able to write 63²⁶ in the form: 

63²⁶ = 16q + r

where q and r are positive integers and r < 16

The trick in solving such questions is to write the base of the given number in terms of the divisor.

The base of 63²⁶ is 63.

So, we have to write 63 in terms of the divisor 16.

The simplest way to do so is:

63 = 16 × 4 - 1

This means, 63²⁶ can be written as follows:

63²⁶ = [(16 × 4) – 1]²⁶

We have expressed 63²⁶ as [(16 × 4) – 1]²⁶. This expression is analogous to the expression (a-b)ⁿ we have discussed above in the explanation of Binomial Theorem.

So, we can observe that when the expression [(16 × 4) – 1]²⁶ is opened up using the Binomial Theorem, every term will have (16 × 4) except the last term

This means, all terms- except the last term- will be divisible by 16

And, what will the last term be?

It will be (-1)²⁶ = 1

So, 63²⁶  = 16E + 1, where 16E represents all terms of [(16 × 4) – 1]²⁶ except the last term

Compare with: 63²⁶ = 16q + r, where r < 16

Thus, we see that Remainder r = 1

Answer: Option (C)

As per the Binomial Theorem, an expression of the form (a-b)ⁿ opens up as follows:

The interesting thing to note here is that the sum of powers of a and b in each term of this expression will be equal to n. For example, if in a term, a is raised to the power n-1, then b will be raised to the power 1 in that term, because (n-1) +1 = n

Similarly, if b is raised to the power n in a term, then a will be raised to the power zero in that term.

Please don’t feel intimidated by the Binomial Theorem. Do you need to know it to crack the GMAT? The answer is No, Not really. We have discussed this expression here just to help you better visualize what’s going on here.

The point that we really want to drive home is that when the expression (a-b)ⁿ is opened up, every term except the last term has a in it.  In fact you can try it out with n = 2 or n = 3 and see it for yourself. 

This means, an expression of the form (a-b)ⁿ can be simplified as ka + (-b)ⁿ , where k is a complicated expression that is outside the scope of the GMAT.

Some students are able to figure out the above simplification intuitively, without any knowledge of the Binomial Theorem, and that is absolutely fine.

Solving the given question using Binomial Theorem

To find the remainder, we should be able to write 63²⁵ in the form: 

• 63²⁵ = 16q + r
  • where q and r are positive integers and r < 16

The trick in solving such questions is to write the base of the given number in terms of the divisor.

The base of 63²⁵ is 63.

So, we have to write 63 in terms of the divisor 16.

The simplest way to do so is:

• 63 = 16 × 4 - 1

This means, 63²⁵ can be written as follows:

• 63²⁵ = [(16 × 4) – 1]²⁵

We have expressed 63²⁵ as

• [(16 × 4) – 1]²⁵.

This expression is analogous to the expression (a-b)ⁿ we have discussed above in the explanation of Binomial Theorem.

So, we can observe that when the expression [(16 × 4) – 1]²⁵ is opened up using the Binomial Theorem, every term will have (16 × 4) except the last term

This means, all terms except the last term will be divisible by 16

And, what will the last term be?

• Last term: (-1)²⁵ = -1

So,

• 63²⁵ = 16E - 1
  • where 16E represents all terms of [(16 × 4) – 1]²⁵ except the last term

Now the remainder for 63²⁵÷16 cannot be equal to – 1 since remainders cannot be negative. 

This expression 63²⁵ = 16E - 1  is exactly parallel to  n = 16k - m. 

So, here, we will apply The Remainder Finding Process B.

We will write m = ad-b where a and b are positive integers and b< d.  The remainder for n ÷ d is equal to b.  

Comparing 63²⁵ = 16E - 1  and  n = 16k - m, we get  m = 1

So, we need to write 1 in the form 16a – b, where b < 16

We can write

• 1 = 16 – 15.

So,

• b = 15

Thus, we can conclude that when 63 raised to the power 25 is divided by 16, the remainder is equal to 15.

Answer: Option (E)

Given: k = 1! + 2! + 3! + . . . + p!

Applying the definition of factorial operation:

n! = 1 × 2 × 3 × 4 × 5 × 6 × 7. . . × (n-2) × (n-1) × n

where n is a positive integer

This implies that:

• 1! = 1
• 2! = 1 x 2
• 3! = 1 x 2 x 3
• 4! = 1 x 2 x 3 x 4
• 5! = 1 x 2 x 3 x 4 x 5
• …
• p! = 1 x 2 x 3 x 4 x 5 x 6 x 7 …x (p-2) x (p-1) x p

From 4! onwards, all terms are divisible by 4.

So, to find the remainder when k is divided by 4, we need to consider only terms till 3!

• 1! + 2! + 3! = 1+ 1 × 2 + 1 × 2 × 3  = 1 + 2 + 6 = 9

The remainder when 9 is divided by 4 is 1.  So when k is divided by 4, the remainder is 1.

Answer: Option (B)

Steps 1 & 2: Understand Question and Draw Inferences

We need to find the remainder when the positive integer n is divided by 2

i.e., we need to find if n is even or odd.

If n is even, then the remainder is 0

If n is odd, then the remainder is 1.

Step 3: Analyze Statement 1

When n is divided by 13, the remainder is 3

• n = 13k + 3

This is not sufficient. n is odd when k is even and even when k is odd. 

Step 4: Analyze Statement 2

n + 2 is a multiple of  7

• n+2 = 7t
• n = 7t -2  = 7(t-1) + 5

This statement is not sufficient. n is even when t-1 is odd and n is odd when t-1 is even

Step 5: Analyze Both Statements Together (if needed)

Inference from statement 1: n = 13k + 3

Inference from statement 2: n = 7t -2 =  7(t-1) + 5

 Inference from statement 1 and statement 2: 13k + 3 = 7(t-1) + 5

• 13k – 7(t-1) = 2
• 13k – 7(t-1) = even

We know:

Odd – Odd = Even

Even – Even = Even

So k and t-1 can be both even or both odd

Hence, we cannot find if k and t is even or odd, and therefore we cannot find if n is even or odd.

Statement 1 and Statement 2 together are not sufficient to answer the question.

 Answer: Option (E)