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GMAT Word Problem Numbers properties, The sum of four consecutive integers is 418. What is the sum of the next four consecutive integers?



Miércoles 25 de julio de 2018, por hurtado claudio

Number Properties N5 2
1.- The sum of four consecutive integers is 418. What is the sum of the next four consecutive integers?

A) 422
B) 431
C) 440
D) 434
E) 424

2.- If 83! has 16 zeroes at the end, how many zeroes will 90! have at the end?

A. 16
B. 17
C. 18
D. 19
E. 20

3.- What is the greatest number of identical groups that can be made out of 130 mens and 91 women if no persons are to be left out? (Two groups are identical whenever the number of man in the two bouquets is equal and the number of women in the two bouquets is equal)

(A) 3
(B) 4
(C) 5
(D) 6
(E) 13

4.- If p is the product of all integers from 25 to 50, inclusive, How many different values can the integer x take, for which 10^x is a factor of p?

(A) 1
(B) 2
(C) 3
(D) 9
(E) 11


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Resp: 1/d, 2/c, 3/e, 4/d

SOLVED

1.- The SUM of two integers consecutive is x, then the SUM of the next two consecutive integers is x + 2².

The SUM of n integers consecutive is p, then the SUM of the next n consecutive integers is p + n²

The sum of four consecutive integers is 418. What is the sum of the next three consecutive integers?
418 + 4² = 434 d)
2.- 83! = 16 zeroes at the end
90! = 83! 84*85*86*87*88*89*90=(16 ceros)*(2*42)*(5*17)*(2*43)*(3*29)*(2*44)*89*(9*10)=
=(10^16)*(2*5)*42*17*2*43*3*29*2*44*89*9*10=(10^16)*10*10*42*17*43*3*29*2*44*89*9=(10^18)*……
90! Increment two zeroes at the end.
16 + 2 = 18 zeroes at the end. c)
3.- Since no persons are to be left out, then the number of groups must be a factor of both 130 and 91. For example, we cannot have 3 groups since we cannot divide 91 women into 3 groups without one women left over.

Only answer choice which is a factor of 91 is E (13).
4.- Simil 2



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