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When p is divided by 7, the remainder is 4. When p is divide
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05 Nov 2017, 01:58
05 Nov 2017, 01:58
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When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?
(A) 2
(B) 4
(C) 5
(D) 11
(E) 18
Kudos for correct solution.
(A) 2
(B) 4
(C) 5
(D) 11
(E) 18
Kudos for correct solution.
Re: When p is divided by 7, the remainder is 4. When p is divide
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05 Nov 2017, 06:56
05 Nov 2017, 06:56
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Here we can notice that if a number divided by 7 leaves a remainder of 4, it means that if we sum 3 to it it becomes divisible by 7. The same holds for the division by 4: since the remainder is 1 when dividing by 4, if we sum 3, we get a number divisible by 4.
Since the number we have to sum to make the number divisible by 7 and 4 is the same, it means that the number is divisible by the product of 7 and 4, i.e. 28. Thus, we have to look for the number of numbers less than 120 which are multiples of 28.
Then, using the formula and given the first multiple equal to 28 and the highest one equal to 112, we get \(\frac{112-28}{28}+1 = 4\).
Answer B
Since the number we have to sum to make the number divisible by 7 and 4 is the same, it means that the number is divisible by the product of 7 and 4, i.e. 28. Thus, we have to look for the number of numbers less than 120 which are multiples of 28.
Then, using the formula and given the first multiple equal to 28 and the highest one equal to 112, we get \(\frac{112-28}{28}+1 = 4\).
Answer B
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Re: When p is divided by 7, the remainder is 4. When p is divide
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09 Jan 2018, 10:11
09 Jan 2018, 10:11
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Expert Reply
Bunuel wrote:
When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?
(A) 2
(B) 4
(C) 5
(D) 11
(E) 18
(A) 2
(B) 4
(C) 5
(D) 11
(E) 18
We have the following two equations:
p = 7Q + 4 where Q is a whole number.
So, p can be values such as 4, 11, 18, 25, …, 7(16) + 4 = 116.
and
p = 4S + 1 where S is a whole number.
So, p can be values such as 1, 5, 13, 14, 21, 25, …, 4(29) + 1 = 117.
We see that 25 is the first matching value of p. Since the LCM of 7 and 4 is 28, the next will be 25 + 28 = 53, and the next will be 53 + 28 = 81, and the final matching value will be 81 + 28 = 109, so there are a total of 4 values of p less than 120 that satisfy the two criteria.
Alternate Solution:
We have
p = 7Q + 4
for some Q and
p = 4S + 1
for some S.
Then, p + 3 = 7Q + 7 = 4S + 4 is divisible by both 7 and 4. Let’s list the numbers less than 120 that are divisible by 7 and 4, i.e., by 28: 0, 28, 56, 84, 112. However, these are values for p + 3; thus, the corresponding values for p are -3, 25, 53, 81, and 109. p can take any value in this list except for -3; therefore, there are four values in total.
Answer: B
