How do I solve ABCD + A + B + C + D = 1991?

 

PLEASE SEE NEW MATERIAL BELOW.

If ABCD is, as it often is, a four digit number with digits A, B, C, D, then write

1000A+100B+10C+D+A+B+C+D=1991

⇒1001A+101B+11C+2D=1991(1)

Clearly A=1 and (1) becomes 101B+11C+2D=990 and then B=9 and we look at 11C+2D=81 from which C=7 and D=2.

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To strengthen the question, the instruction solve is not enough. We need to know where the variables live. As stated, we can only assume the worst and that suggests the implied question,

Find complex numbers a, b, c, d such that …

Then the only answer is of the form d=1991−a−b−cabc

which represents an infinite number of quartets. Just choose your favourite numbers for a, b and c. Don’t forget that you have to choose all of them.

To make a question more meaningful, easier or tasteful, we often imagine words which when included remove, at least in our minds, any ambiguity. That is what I did rephrasing the question in my mind as

Solve the following cryptarithm.

Such alterations are acceptable if there is a precedent. A quick check on Quora reveals over 247 examples where abc etc is not considered a product. Over 247 because that is when I lost count - quite true that I have no life. So considering it a cryptarithm is legit.

It is often done, that letters at the beginning of the alphabet are used to represent integers. Then the following words might be in our minds:

Find integers a, b, c, d such that …

However, even with just a=0, b+c+d=1991 has an infinite number of solutions all of the form d=1991-b-c. Still not pleasing to present.

We could restrict our variable home to non-zero integers but there would still be an infinite number of solutions, perhaps characterized by those values of a, b, c so that 1991−a−b−cabc

is an integer.

This leads us to our final replacements for solve:

 _______________

Find all positive integers, a, b, c, d, such that…

We include “all” so that no one just offers a,a,a,1991

and leaves.

Sometimes, a precedent is noted in writing a solution. Here, I will say

Without loss of generality, assume that a≤b≤c≤d

but be aware that there may be up to 24 times as many solutions as shown. Since a is the smallest, a4+4a≤1991 and so a≤6 and so is of {1, 2, 3, 4, 5, 6} and we handle these separately as cases. With a as listed, look at ab3+a+3b≤1991

and see that b has maximum values {12, 9, 8, 7, 7, 6} respectively. Only 34 cases to consider. Let us be at it.

If a=1

and b=1, abcd+a+b+c+d=1991

(the equation) , becomes

cd+c+d=1989⇒(c+1)(d+1)=1990

c+1

and d+1 are make a factor pair of 1990, smaller one first. So (a,b,c,d)=(1,1,1,994),(1,1,4,397),(1,1,9,198)

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If a=1 and b=2, the equation is 2cd+c+d=1988

and so (2c+1)(2d+1)=3977.

2c+1

and 2d+1 are make a factor pair of 3977 and so (a,b,2c+1,2d+1)=(1,2,41,97) and then (a,b,c,d)=(1,2,20,48)

.

If a=1 and b=3, the equation is 3cd+c+d=1987

and so (3c+1)(3d+1)=5962. 3c+1 and 3d+1 make a factor pair of 5962 and so (a,b,3c+1,3d+1)=(1,3,11,542) and (1,3,22,271) giving (a,b,c,d)=1,3,7,90) since (1,3,11,542)

does not cooperate.

If a=1 and b=4, the equation is 4cd+c+d=1986

and so (4c+1)(4d+1)=7945. 4c+1 and 4d+1 make a factor pair of 7945 and so (a,b,4c+1,4d+1)=(1,4,5,1589),(1,4,7,1135) and (1,4,35,227)

none of which cooperate.

Let’s call everything after If a=1 and b=4, stuff.

If a=1 and b=5 or 6, stuff gives no solutions

If a≠1

stuff gives no solutions.

Here we have covered 4 interpretations all of which have precedents and are valid.

NOTE: If there is an easy way to see that a=1 we would be on the verge of elegance.

#GK