Geometry Puzzle – Critical Thinking 20

Challenge:

In how many 3-digit numbers the sum of all digits is 7?

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First, let’s find all three digits that add up to 7.
700, 610, 511, 520, 430, 421, 322, 133
For 7, 0 and 0, we can write only one three digit number which is 700.
For 6, 1 and 0, we can find 106, 160, 601, 610,
For 5, 1, and 1: 115, 151, 511
For 5, 2, and 0: 205, 250, 502, 520
For 4, 3, and 0: 304, 340, 403, 430
For 4, 2, and 1: 124, 142, 214, 241, 412, 421
For 3, 2, and 2: 223, 232, 322
For 1, 3, and 3: 133, 313, 331
There are 28 three digits numbers that the sum of their digits is 7.

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Frequently Asked Questions

How does stars and bars apply here?

You’re distributing 7 units (the digit-sum) across 3 slots (the digits) with nonnegative integers — that’s \( \binom{n+k-1}{k-1} \) where \( n=7 \) and \( k=3 \). Initial count: \( \binom{9}{2} = 36 \). Then subtract the cases where the hundreds digit is 0 to honor the three-digit constraint.

Where does the substitution \( a-prime= a-1 \) come from?

It turns a strictly-positive variable (\( a \geq 1 \)) into a nonnegative one (\( a-prime\geq 0 \)), which is the form stars and bars handles directly. After substitution, \( a + b + c = 7 \) becomes \( a-prime+ b + c = 6 \). Count \( \binom{6+2}{2} = \binom{8}{2} = 28 \).

Why is no upper-bound constraint needed?

Digits cap at 9, but \( a + b + c = 7 \) means no digit can possibly exceed 7. So we don’t have to subtract any \”digit too big\” cases. If the sum were, say, 20, we’d also have to subtract scenarios where a digit overshot 9.

Can I list a few of the 28 numbers as a sanity check?

Sure. \( 700, 610, 601, 520, 511, 502, 430, 421, 412, 403, \ldots, 106, 115, 124, 133, 142, 151, 160 \). Pair each \((a, b, c)\) with \( a + b + c = 7 \) and \( a \geq 1 \) — you’ll list exactly 28.

What if I want digits adding to a different total?

Same recipe. For \( a + b + c = N \) with \( a \geq 1 \) and \( a, b, c \leq 9 \): substitute \( a’ = a – 1 \), get \( a’ + b + c = N – 1 \), apply stars and bars \( \binom{N+1}{2} \), then subtract overcount for any digit exceeding 9 using inclusion-exclusion.

How is this puzzle related to combinatorics in general?

It’s a classic intro to integer compositions and partitions — counting ways to write a positive integer as an ordered sum. Stars and bars is the workhorse tool, and these digit-sum puzzles are a friendly way to introduce it.

What’s an alternative way to count without stars and bars?

Loop through \( a \) from 1 to 7. For each \( a \), count nonnegative \((b, c)\) with \( b + c = 7 – a \) — that’s \( 7 – a + 1 = 8 – a \) pairs. Sum: \( \sum_{a=1}^{7}(8-a) = 7+6+5+4+3+2+1 = 28 \). Same answer.

Why exclude the hundreds digit being 0?

Because then the number wouldn’t be a 3-digit number — it would have a leading zero, making it effectively 2-digit or shorter. The puzzle explicitly asks about 3-digit numbers, so the hundreds digit must be 1 or larger.

What grade level can solve this?

The brute-force counting works for late elementary or middle school. The clean stars-and-bars solution suits Algebra II, Precalculus, or any combinatorics unit. Either path lands on 28.

Where else does this kind of counting show up?

Anytime you distribute identical objects into distinct bins: distributing candy among children, splitting tasks across servers, counting integer solutions to constrained equations. The stars-and-bars formula is one of the most reusable tools in combinatorics.

Related Lessons You May Like

• How to solve permutations and combinations
• How to find the probability of an event
• How to solve multi-step word problems
• How to add and subtract multi-digit numbers
• Place value up to thousands

If your student likes puzzles like this one, Geometry for Beginners works the same kinds of relationships inside a full geometry curriculum. For the algebra you'll lean on for setting up the equations, Pre-Algebra for Beginners fills in the foundations gently.