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TMUA 2018 · Paper 2 · Question 7 of 20

TMUA 2018 Paper 2 Question 7

Number and divisibility — Overlap of two arithmetic progressions · remainders. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 2Number and divisibilityOverlap of two arithmetic progressions · remainders7 options
Sequence 1 is an arithmetic progression with first term 11 and common difference 3.

Sequence 2 is an arithmetic progression with first term 2 and common difference 5.

Some numbers that appear in Sequence 1 also appear in Sequence 2. Let N be the 20th such number.

What is the remainder when N is divided by 7?

  1. A0
  2. B1
  3. C2
  4. D3
  5. E4
  6. F5
  7. G6
Show the answer and worked solution
answer · B
  1. A0
  2. B1
  3. C2
  4. D3
  5. E4
  6. F5
  7. G6
Sequence 1 consists of the numbers that are 2 more than a multiple of 3 and at least 11; Sequence 2 consists of the numbers that are 2 more than a multiple of 5 and at least 2. A number in both is 2 more than a multiple of 15, so the shared numbers are 17,  32,  47,  … — note that 2 itself is too small for Sequence 1. The kth is 15k + 2, so N = 15× 20 + 2 = 302. Since 302 = 7× 43 + 1, the remainder is 1.