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Is the product of four consecutive integers a multiple of 120?
(1) None of the four digits is a multiple of 10.
(2) The sum of the four integers is a multiple of 10.
(1) None of the four digits is a multiple of 10.
(2) The sum of the four integers is a multiple of 10.
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B.
interesting question
(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.
Therefore, B is the correct answer.
interesting question
(1) is not sufficient. Note: 120=2*2*2*3*5, for example 23*24*25*26 is a multiple of 120, or 120*121*122*123.
(2) is sufficient.
We know that the last digits of the four numbers must be either 1, 2, 3 and 4 (1+2+3+4=10) or 6, 7, 8, 9 , these numbers constructed in that way
(i. e. k*10+1, k*10+2, K*10+3, k*10+4, k integer)
are not divisible by 5 (analogously for 6, 7, 8, 9). Hence, We know that these numbers cannot be multiples of 120.
Therefore, B is the correct answer.
