document.write( "Question 1070377: In the equation, N × U × (M + B + E + R) = 33, each letter stands for a different digit.
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Algebra.Com's Answer #685517 by AnlytcPhil(1806) $\"\"$ $\"About$

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N × U × (M + B + E + R) = 33
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document.write( "The only factors of 33 are 1,3,11,33\r\n" );
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document.write( "N × U must be one of the factors and M + B + E + R must be another.\r\n" );
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document.write( "N × U can only be 3, \r\n" );
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document.write( "for if it were 1, M + B + E + R would have to be 33 -- impossible!\r\n" );
document.write( "and if it were 11, M + B + E + R would have to be 3 -- also impossible! \r\n" );
document.write( "and if it were 33, M + B + E + R would have to be 1 -- also impossible!\r\n" );
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document.write( "So N and U are 1 and 3, and M + B + E + R = 11\r\n" );
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document.write( "So we must pick four digits from 0,2,4,5,6,7,8,9\r\n" );
document.write( "that have sum 11.\r\n" );
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document.write( "The only ones that have sum 11 are the smallest four, \r\n" );
document.write( "0 + 2 + 4 + 5 = 11\r\n" );
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document.write( "So the first way is\r\n" );
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document.write( "1 × 3 × (0 + 2 + 4 + 5) = 33\r\n" );
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document.write( "The 1 and 3 can be arranged in 2! = 2 ways\r\n" );
document.write( "The 0,2,4, and 5 can be arranged in 4! = 24 ways\r\n" );
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document.write( "Answer:   2! × 4! = 2 × 24 = 48\r\n" );
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document.write( "Edwin

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