What is the largest number which when divided into 140 and 121will leave a remainder of 12 in each case?

Answer 1

There are several ways to interpret this question. We begin with the assumption that the question is being asked within the contents of a typical secondary school mathematics class.

Answer 1:

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I assume we are looking for the largest integer divisor that produces 12 as a remainder. To obtain the answer we calculate each number minus the remainder of 12, and then derive their largest common factor. If that common factor is greater than 12 it is a proper divisor for that remainder and you have the answer. If not, there is no solution to the question and therefore question cannot be answered.

140 - 12 = 128 (2 raised to the 7th power)

121 - 12 = 109 (a Prime number)

Since the greatest common factor of these two numbers is one, there is no proper answer to this question, and this is, in essence, a "trick question." Properly phrased, it should have asked if there was such a number and if so, what would it be and if not, to provide a proof that none exists.

Because there is not a proper answer as interpreted above, we look at the question more technically, i.e., we consider the actual statement rather than what we might have initially assumed was intended. This leads to several other possible solutions.

Answer 2

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Let X be the number we are looking for (the largest divisor of 140 and 121 that leaves 12 as a remainder). Because the remainder is an integer, we can then express the requirements as follows:

For some integers Y and Z,

[Equation 1]

140=XZ+12, (usually we would require 12<X, but to make the solution more general, we don't make this assumption)

and

[Equation 2]

121=XY+12, (again, we would usually require 12<X, but here we don't make this assumption)

Subtracting the second equation from the first, we obtain

19=XZ-XY

19=X(Z-Y) (distributive law)

X=19/(Z-Y) (solving for X)

Rewriting Equation 2 using this value for X, we obtain

121=(19/(Z-Y))Y+12

We simplify this expression

109=19Y/(Z-Y)

109Z-109Y=19Y

109Z=128Y

Since 109 is prime and 128 is a power of 2, they are relatively prime and Z must contain 128 as a factor and Y must contain 109 as a factor.

One possibility is

Z=128

and

Y=109

So that X= 19/(Z-Y)=19/19=1

Thus

140=1(128)+12 and

121=1(109)+12.

So although we don't usually consider 140 divided by 1 to be a quotient of 128 with a remainder of 12 and 121 divided by 1 to be a quotient of 109 with a remainder of 12, this divisor of 1 does technically satisfy the requirements of the question (regardless of what the proposer intended and the way such questions are usually interpreted in typical high school math classes).

Is there a larger number? If so, we must have

X= 19/(Z-Y)>1

We previously assumed

140=XZ+12

128=XZ so that

128/Z=X>1

128/Z>1

128>Z

But such a Z cannot contain 128 as a factor, so no such Z exists. Thus there are no other values of X that will satisfy the requirements of the question and 1 is the largest that does satisfy the requirements.

So the answer is 1 for this interpretation.

Answer 3

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This solution looks more carefully at the possible interpretations of the phrase "leaves a remainder." This is usually interpreted as something left over from the usual operation of division. However, it can also be interpreted as the phrase that would be used when the division operation is looking for a quotient which acts as a ceiling function rather than a floor function. For example, we might be looking for ways to divide up an area of 140 square inches into a number of subsections with equal areas but the operation needs to be within 12 of a complete subsection, in other words, "leave a remainder of 12" in order to complete the next largest whole subsection. In this case, the remainder specifies the remaining unfulfilled requirements in order to complete another subsection (perhaps because this amount is available from some other source) rather than the partially filled amount of another subsection provided by the operation itself. In this interpretation, the equations become

[Equation 3]

140 +12 = XZ,

and

[Equation 4]

121 + 12 = XY

These reduce to

152 = XZ and

133 = XY.

Looking at the prime decomposition of 133 and 152, we have

(19)(2)(2)(2) = XZ

(19)(7) = XY

Assuming X , Y, and Z are each > 1, and since the only possible common factor of the two numeric values is 19, that must be the value of X. Thus we have

(140+12)/19 = 8 (exactly)

and

(121+12)/19 = 7 (exactly)

So the answer is 19 for this interpretation.

Answer 4

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It is possible that the question was mistyped when it was entered. For example, a single keypad typographical error (typing "1" instead of "4", which are two adjacent keys) would imply that the following was intended: "What is the largest number which when divided into 140 and 124 will leave a remainder of 12 in each case?"

The solution would then proceed as follows.

140-12=128=16*8

124-12=112=16*7

So the greatest common factor is 16 and we have

140/16 is 8 with a remainder of 12 and

124/16 is 7 with a remainder of 12.

So the answer is 16 for this interpretation of the question.