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Re: Hundred's digit of sum of numbers
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26 Jan 2012, 12:05
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Both a, b, and c are 3-digits integers, where a=b+c. Is the hundreds' digit of number a equal to sum of that of b and c?
Hundreds digit of \(a\) will equal to the sum of hundreds digits of \(b\) and \(c\) if there is no carried over 1 from the sum of the tens digits of \(b\) and \(c\). To illustrate:
149=b 249=b --- 398=a As you can see there is no carried over 1 from the sum of 4+4+1=9 (1 is carried over from the sum of units digits) and thus hundreds digit of \(a\) equal to the sum of hundreds digits of \(b\) and \(c\): 3=1+2;
But if: 170=b 240=b --- 410=a As you can see there is a carried over 1 from the sum of 7+4=11 and thus hundreds digit of \(a\) doesn't equal to the sum of hundreds digits of \(b\) and \(c\): \(4\neq{1+2}.\)
(1) Tens' digit of a=tens' digit of b+tens' digit of c --> there can not be carried over 1 from this sum as it must be less than 10 to equal to tens' digit of \(a\). Sufficient.
(2) Units' digit of a=units' digit of b+units' digit of c. Clearly insufficient as no info about tens digits.
If a, b, and c are 3-digit positive integers, where a=b+c,
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Updated on: 25 Jun 2013, 05:09
If a, b, and c are 3-digit positive integers, where a=b+c, is the hundreds' digit of a equal to the hundreds' digit of b plus the hundreds' digit of c?
(1) The tens' digit of a is equal to the tens digit of b plus the tens' digit of c. (2) The units' digit of a is equal to the units' digit of b plus the units' digit of c.
Re: If a, b, and c are 3-digit positive integers, where a=b+c,
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25 Jun 2013, 17:26
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stunn3r wrote:
If a, b, and c are 3-digit positive integers, where a=b+c, is the hundreds' digit of a equal to the hundreds' digit of b plus the hundreds' digit of c? (1) The tens' digit of a is equal to the tens digit of b plus the tens' digit of c. (2) The units' digit of a is equal to the units' digit of b plus the units' digit of c.
What is the source of this question?
The sum of the hundred's digits of a & b will equal the hundred's digit of c if nothing "carries" from the tens column. 123 + 351 = 474 123 + 358 = 481 Both of those lead to "yes" answer to the prompt question. In order for statement #1 to be true, it must be true that nothing carries from the ones column to the tens column (only the first of the two example addition statements satisfies this) --- thus, statement #1 automatically implies statement #2. That's why statement #1, by itself, is sufficient --- it already completely contains the information in statement #2. That's why (A) has to be the answer.
In other words, I am claiming that it is utterly impossible to find numbers that satisfy both the prompt condition and statement #1 but not statement #2. In order for (C) to be the answer, there would have to be a case in which you could add two three digit numbers which met statement #1, did not meet statement #2, and the three-digit sum did not work out neatly in the hundreds column. You would have to able to find such a case, and I claim that finding such a case is impossible. That's why (C) can't be the answer.
Re: If a, b, and c are 3-digit positive integers, where a=b+c,
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26 Jun 2013, 02:00
mikemcgarry wrote:
stunn3r wrote:
If a, b, and c are 3-digit positive integers, where a=b+c, is the hundreds' digit of a equal to the hundreds' digit of b plus the hundreds' digit of c? (1) The tens' digit of a is equal to the tens digit of b plus the tens' digit of c. (2) The units' digit of a is equal to the units' digit of b plus the units' digit of c.
What is the source of this question?
The sum of the hundred's digits of a & b will equal the hundred's digit of c if nothing "carries" from the tens column. 123 + 351 = 474 123 + 358 = 481 Both of those lead to "yes" answer to the prompt question. In order for statement #1 to be true, it must be true that nothing carries from the ones column to the tens column (only the first of the two example addition statements satisfies this) --- thus, statement #1 automatically implies statement #2. That's why statement #1, by itself, is sufficient --- it already completely contains the information in statement #2. That's why (A) has to be the answer.
In other words, I am claiming that it is utterly impossible to find numbers that satisfy both the prompt condition and statement #1 but not statement #2. In order for (C) to be the answer, there would have to be a case in which you could add two three digit numbers which met statement #1, did not meet statement #2, and the three-digit sum did not work out neatly in the hundreds column. You would have to able to find such a case, and I claim that finding such a case is impossible. That's why (C) can't be the answer.
Does all this make sense? Mike
hmm .. true .. Thanks mike .. I figured it out when I retried the question but your explanation made it more clear.
I've this hard copy of 200 mixed bag questions .. these are good concept building and "makes you scratch your head a little" type questions. I am posting all which I think would give something to learn to forum members and the questions which I get wrong I did first 40 of this set yesterday and posted around 7-8 questions. will post more, keep answering. _________________
Re: If a, b, and c are 3-digit positive integers, where a=b+c,
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01 Jul 2013, 13:06
stunn3r wrote:
If a, b, and c are 3-digit positive integers, where a=b+c, is the hundreds' digit of a equal to the hundreds' digit of b plus the hundreds' digit of c?
(1) The tens' digit of a is equal to the tens digit of b plus the tens' digit of c. (2) The units' digit of a is equal to the units' digit of b plus the units' digit of c.
Re: Both a, b, and c are 3-digits integers, where a=b+c. Is the
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17 Mar 2019, 07:42
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61% (01:44) correct 
and the questions which I get wrong 
