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22 Jul 2022, 08:00
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D
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Difficulty:
55%
(hard)
Question Stats:
60% (02:02) correct
40%
(02:26)
wrong
based on 191
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Is the sum of four consecutive integers a multiple of 10 ?
(1) The product of the four integers is a multiple of 120.
(2) The product of the units digits of the four integers is a multiple of 120.
(1) The product of the four integers is a multiple of 120.
(2) The product of the units digits of the four integers is a multiple of 120.
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22 Jul 2022, 08:22
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Bunuel
Assume the integers be I, I+1, I+2, I+3. Question is 4I+7 is a multiple of 10 or not
(1) The product of the four integers is a multiple of 120: 120 is 2*2*2*3*5, or 2*3*4*5. So for any value, summation will not be a multiple of 10. Sufficient
(2) The product of the units digits of the four integers is a multiple of 120: numbers can be 12,13,14,15; 22,23,24,25; 32,33,34,35 ..... Again for any value, summation will not be a multiple of 10. Sufficient
D is the correct option
22 Jul 2022, 09:00
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Asked: Is the sum of four consecutive integers a multiple of 10 ?
Let the consecutive integers be x, x+1, x+2 & x+3.
Is (x+x+1+x+2+x+3) = 4x + 6 is a multiple of 10 ?
(1) The product of the four integers is a multiple of 120.
x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x+2) = (t-1)(t+1) = t^2 - 1; where t = x^2 + 3x +1
t^2 - 1 = 120k; t^2 - 120k - 1 = 0; \(t = \sqrt{120k+1}\) or \(t = -\sqrt{120k+1}\)
If k =1; t^2 = 121; For other values of k, t is NOT an integer. Therefore, x is NOT an integer.
t=x^2 + 3x +1 = 11; x^2 + 3x -10 = 0; (x+5)(x-2) = 0; x = 2 or x = -5
t=x^2 + 3x +1 = -11; x^2 + 3x +12 = 0; x is NOT an integer
Consecutive integers = {-5,-4,-3,-2} or {2,3,4,5}
Sum of 4 integers = (2+3+4+5) = 14 or -(2+3+4+5) = -14;
Sum of 4 integers is NOT a multiple of 10.
SUFFICIENT
(2) The product of the units digits of the four integers is a multiple of 120.
2*3*4*5 = 120 = 120*1 : Sum of unit digits = 2+3+4+5 = 15: NOT a multiple of 10
3*4*5*6 = 360 =120*3: Sum of unit digits = 3+4+5+6 = 18: NOT a multiple of 10
4*5*6*7 = 840 = 120*7: Sum of unit digits =4+5+6+7 = 22: NOT a multiple of 10
5*6*7*8 = 1680 = 120*14: Sum of unit digits = 5+6+7+8 = 26: NOT a multiple of 10
6*7*8*9 is NOT a multiple of 120
7*8*9*0 = 0 = 120*0: Sum of unit digits =7+8+9+0 = 24: NOT a multiple of 10
SUFFICIENT
IMO D
Let the consecutive integers be x, x+1, x+2 & x+3.
Is (x+x+1+x+2+x+3) = 4x + 6 is a multiple of 10 ?
(1) The product of the four integers is a multiple of 120.
x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x+2) = (t-1)(t+1) = t^2 - 1; where t = x^2 + 3x +1
t^2 - 1 = 120k; t^2 - 120k - 1 = 0; \(t = \sqrt{120k+1}\) or \(t = -\sqrt{120k+1}\)
If k =1; t^2 = 121; For other values of k, t is NOT an integer. Therefore, x is NOT an integer.
t=x^2 + 3x +1 = 11; x^2 + 3x -10 = 0; (x+5)(x-2) = 0; x = 2 or x = -5
t=x^2 + 3x +1 = -11; x^2 + 3x +12 = 0; x is NOT an integer
Consecutive integers = {-5,-4,-3,-2} or {2,3,4,5}
Sum of 4 integers = (2+3+4+5) = 14 or -(2+3+4+5) = -14;
Sum of 4 integers is NOT a multiple of 10.
SUFFICIENT
(2) The product of the units digits of the four integers is a multiple of 120.
2*3*4*5 = 120 = 120*1 : Sum of unit digits = 2+3+4+5 = 15: NOT a multiple of 10
3*4*5*6 = 360 =120*3: Sum of unit digits = 3+4+5+6 = 18: NOT a multiple of 10
4*5*6*7 = 840 = 120*7: Sum of unit digits =4+5+6+7 = 22: NOT a multiple of 10
5*6*7*8 = 1680 = 120*14: Sum of unit digits = 5+6+7+8 = 26: NOT a multiple of 10
6*7*8*9 is NOT a multiple of 120
7*8*9*0 = 0 = 120*0: Sum of unit digits =7+8+9+0 = 24: NOT a multiple of 10
SUFFICIENT
IMO D
