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# What is the remainder when the 4-digit positive number, 1,000a + 100b

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Math Expert
Joined: 02 Sep 2009
Posts: 58365
What is the remainder when the 4-digit positive number, 1,000a + 100b  [#permalink]

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29 Mar 2018, 00:01
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Difficulty:

35% (medium)

Question Stats:

66% (01:15) correct 34% (01:34) wrong based on 46 sessions

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What is the remainder when the 4-digit positive number, 1,000a + 100b + 10c + d, is divided by 3?

(1) a + b + c + d = 17
(2) (1,000a + 100b + 10c + d) - 2 is divisible by 3.

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Re: What is the remainder when the 4-digit positive number, 1,000a + 100b  [#permalink]

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29 Mar 2018, 00:21
Bunuel wrote:
What is the remainder when the 4-digit positive number, 1,000a + 100b + 10c + d, is divided by 3?

(1) a + b + c + d = 17
(2) (1,000a + 100b + 10c + d) - 2 is divisible by 3.

Remainder-type questions can often be solved based on number properties, without making explicit calculations.
This is a Logical approach.

A number is diviisible by 3 if the sum of its digit is divisible by 3.

(1) Since the digits of our numbers sum to 17, they are one less than a number whose digits sum to 18 (or two more than a number whose digits sum to 15). That is, the closest multiple of 3 is 2 less than our number meaning it has a remainder of 2 when divided by 3.
Sufficient.

(2) This directly tells us that the remainder of our number divided by 3 is 2.
Sufficient.

(D) is our answer.
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Re: What is the remainder when the 4-digit positive number, 1,000a + 100b   [#permalink] 29 Mar 2018, 00:21
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# What is the remainder when the 4-digit positive number, 1,000a + 100b

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