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3 votes
Cog counted down from 2021 by subtracting 7 repeatedly. How many positive multiples of 3 did Cog count?

1 Answer

7 votes

Answer:

96

Explanation:

There are 2021 numbers and subtracting 7 each time, we will be left with 2021/7 ≈ 289 numbers

2021/3 = remainder 2

⇒2021 = 3x + 2

2021 - 7 = 3x + 2 - 7

= 3x - 5

= 3x - 3 - 2

= 3(x - 1) - 2

adding and sub 3,

= 3(x - 1) - 2 + 3 - 3

= 3(x - 1 - 1) + 1

=3(x - 2) + 1

The remainder is 1 and hence not divisible by 3

2021 - 7 - 7 = 3(x - 1) - 2 - 7

= 3(x - 1) - 9

= 3(x - 1 - 3)

= 3 (x - 4)

Remainderis 0 and hence divisible by 3

So, every 3rd number in the 289 numbers is divisible by 3

⇒ 289/3 ≈ 96

Cog counted 96 numbers

answered
User Andreas Profous
by
8.6k points
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