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## GRE - Quantitative Reasoning - Number Theory

### When m is divided by 11, the remainder is r. (more)

m = 1032  2

When m is divided by 11, the remainder is r.

Quantity A: r

Quantity B: 3

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

m = 1032 + 2 = (11 - 1)32 + 2

expanding using binomial

1132 + b 1131 + . . .  + k11 + 1 + 2 = multiple of 11 + 3

remainder will be 3

### For each integer n > 1, let A(n) denote the sum of the integers from 1 to n. For example, A(100) = 1 + 2 + 3 + â€¦ + 100 = 5,050. What is the value of A(200)? (more)

For each integer n > 1, let A(n) denote the sum of the integers from 1 to n. For example, A(100) = 1 2 3 … 100 = 5,050. What is the value of A(200)?

A. 10,100
B. 15,050
C. 15,150
D. 20,100
E. 21,500

A(200) = 1 + 2 + . . . + 200

Sum of consecutive numbers starting from 1 = n(n+1)/2 = 200*201/ 2 = 20100

D is correct

### If j and k are integers and j â€” k is even, which of the following must be even? (more)

If j and k are integers and j — k is even, which of the following must be even?

A) k

B) jk

C) j 2k

D) jk j

E) jk — 2j

Given j - k is even

=> either both j and k are odd or both are even

Option A - Incorrect because k will be odd if j is odd and k will be even if j is even

Option B. Incorrect because if j,k are odd, the product is odd, if they are even product is even

Option C. Incorrect j + 2k = j - k + 3k = even + 3k dependent on value of k

Option D. jk + j = j ( k + 1), if j is odd, k will be odd -> k + 1 will be even, hence j (k+1) will be even

If j is even j (k+1) will always be even

Correct

Correct option D.