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Exercise 5.1 · Q3

Q.The number of two digit numbers divisible by 6 are:

(a) 24
(b) 14
(c) 15
(d) 20
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✓ Free question

Two-digit multiples of 66 form an A.P. from the smallest (1212) to the largest (9696) two-digit multiple; count the terms with the standard formula.

For an A.P.:

n=an−a1d+1n = \frac{a_n-a_1}{d}+1

  1. Find the first two-digit multiple of 66: 6×2=126\times2=12 (since 6×1=66\times1=6 is only one digit). So a1=12a_1=12.
  2. Find the last two-digit multiple of 66: 6×16=966\times16=96 (since 6×17=1026\times17=102 is three digits). So an=96a_n=96.
  3. Common difference: d=6d=6 (consecutive multiples of 66).
  4. Apply the term-count formula:

n=96−126+1=846+1=14+1=15n = \frac{96-12}{6}+1 = \frac{84}{6}+1 = 14+1=15

  1. Self-check: the multiples are 12,18,24,…,9612,18,24,\ldots,96; dividing each by 66 gives 2,3,4,…,162,3,4,\ldots,16 — that's 16−2+1=1516-2+1=15 integers. ✓
  2. Verify against options: (a) 2424 ✗, (b) 1414 ✗, (c) 1515 ✓, (d) 2020 ✗.
✓Final answer

There are 1515 two-digit numbers divisible by 66 — option (c)

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