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In the former Soviet Union, rubles came in denominations of
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27 Jul 2014, 07:50
27 Jul 2014, 07:50
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In the former Soviet Union, rubles came in denominations of 1, 3, 5, 10, 25, 50, and 100. Boris and Natasha are at the bank to change an enormous pile of 50 and 100 ruble notes, but the teller is out of 10s and is forced to give them their change in nothing but 1, 3, 5, and 25 ruble notes. Did the teller give Boris and Natasha the correct change?
(1) The teller gave Boris and Natasha 2013 notes.
(2) The teller did not give Boris and Natasha any 3 ruble notes.
(1) The teller gave Boris and Natasha 2013 notes.
(2) The teller did not give Boris and Natasha any 3 ruble notes.
Re: In the former Soviet Union, rubles came in denominations of
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27 Jul 2014, 16:48
27 Jul 2014, 16:48
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In the former Soviet Union, rubles came in denominations of 1, 3, 5, 10, 25, 50, and 100. Boris and Natasha are at the bank to change an enormous pile of 50 and 100 ruble notes, but the teller is out of 10s and is forced to give them their change in nothing but 1, 3, 5, and 25 ruble notes. Did the teller give Boris and Natasha the correct change?
Notice that since Boris and Natasha have a pile of 50 and 100 ruble notes, then the entire amount is even: 50x + 100y = even + even = even..
Say the teller gave them a, b, c, and d 1, 3, 5, and 25 ruble notes, respectively. So, a + 3b + 5c + 25d must be even. This could happen in the following cases:
(i) all four terms are even, which would mean that each a, b, c, and d must be even. In this case the sum of the notes = a + b + c + d = even + even + even + even = even.
(ii) two terms are even and the remaining two terms are odd, which would mean that two out of a, b, c, and d are even and the remaining two are odd. In this case the sum of the notes = a + b + c + d = even + even + odd + odd = even.
(iii) all four terms are odd, which would mean that each a, b, c, and d must be odd. In this case the sum of the notes = a + b + c + d = odd + odd + odd + odd = even.
(1) The teller gave Boris and Natasha 2013 notes. The sum of the notes is odd, but we know that it must be even. So, something went wrong, the teller did not give Boris and Natasha the correct change. Sufficient.
(2) The teller did not give Boris and Natasha any 3 ruble notes. b = 0 = even. Clearly insufficient.
Answer: A.
Hope it's clear.
Notice that since Boris and Natasha have a pile of 50 and 100 ruble notes, then the entire amount is even: 50x + 100y = even + even = even..
Say the teller gave them a, b, c, and d 1, 3, 5, and 25 ruble notes, respectively. So, a + 3b + 5c + 25d must be even. This could happen in the following cases:
(i) all four terms are even, which would mean that each a, b, c, and d must be even. In this case the sum of the notes = a + b + c + d = even + even + even + even = even.
(ii) two terms are even and the remaining two terms are odd, which would mean that two out of a, b, c, and d are even and the remaining two are odd. In this case the sum of the notes = a + b + c + d = even + even + odd + odd = even.
(iii) all four terms are odd, which would mean that each a, b, c, and d must be odd. In this case the sum of the notes = a + b + c + d = odd + odd + odd + odd = even.
(1) The teller gave Boris and Natasha 2013 notes. The sum of the notes is odd, but we know that it must be even. So, something went wrong, the teller did not give Boris and Natasha the correct change. Sufficient.
(2) The teller did not give Boris and Natasha any 3 ruble notes. b = 0 = even. Clearly insufficient.
Answer: A.
Hope it's clear.
General Discussion
In the former Soviet Union, rubles came in denominations of
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Updated on: 18 Apr 2015, 03:03
Updated on: 18 Apr 2015, 03:03
Bunuel wrote:
In the former Soviet Union, rubles came in denominations of 1, 3, 5, 10, 25, 50, and 100. Boris and Natasha are at the bank to change an enormous pile of 50 and 100 ruble notes, but the teller is out of 10s and is forced to give them their change in nothing but 1, 3, 5, and 25 ruble notes. Did the teller give Boris and Natasha the correct change?
Notice that since Boris and Natasha have a pile of 50 and 100 ruble notes, then the entire amount is even: 50x + 100y = even + even = even..
Say the teller gave them a, b, c, and d 1, 3, 5, and 25 ruble notes, respectively. So, a + 3b + 5c + 25d must be even. This could happen in the following cases:
(i) all four terms are even, which would mean that each a, b, c, and d must be even. In this case the sum of the notes = a + b + c + d = even + even + even + even = even.
(ii) two terms are even and the remaining two terms are odd, which would mean that two out of a, b, c, and d are even and the remaining two are odd. In this case the sum of the notes = a + b + c + d = even + even + odd + odd = even.
(iii) all four terms are odd, which would mean that each a, b, c, and d must be odd. In this case the sum of the notes = a + b + c + d = odd + odd + odd + odd = even.
(1) The teller gave Boris and Natasha 2013 notes. The sum of the notes is odd, but we know that it must be even. So, something went wrong, the teller did not give Boris and Natasha the correct change. Sufficient.
(2) The teller did not give Boris and Natasha any 3 ruble notes. b = 0 = even. Clearly insufficient.
Answer: A.
Hope it's clear.
Notice that since Boris and Natasha have a pile of 50 and 100 ruble notes, then the entire amount is even: 50x + 100y = even + even = even..
Say the teller gave them a, b, c, and d 1, 3, 5, and 25 ruble notes, respectively. So, a + 3b + 5c + 25d must be even. This could happen in the following cases:
(i) all four terms are even, which would mean that each a, b, c, and d must be even. In this case the sum of the notes = a + b + c + d = even + even + even + even = even.
(ii) two terms are even and the remaining two terms are odd, which would mean that two out of a, b, c, and d are even and the remaining two are odd. In this case the sum of the notes = a + b + c + d = even + even + odd + odd = even.
(iii) all four terms are odd, which would mean that each a, b, c, and d must be odd. In this case the sum of the notes = a + b + c + d = odd + odd + odd + odd = even.
(1) The teller gave Boris and Natasha 2013 notes. The sum of the notes is odd, but we know that it must be even. So, something went wrong, the teller did not give Boris and Natasha the correct change. Sufficient.
(2) The teller did not give Boris and Natasha any 3 ruble notes. b = 0 = even. Clearly insufficient.
Answer: A.
Hope it's clear.
Bunuel,
I did not understand.Why we have not taken this scenario;
E+E+E+O=ODD?
