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Algebra Difficulty 2.6 Junior Find the answer

In the addition problem shown, m,n,pm, n, p, and qq represent positive digits. What is the value of m+n+p+qm+n+p+q?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

From the ones column, we see that 3+2+q3 + 2 + q must have a ones digit of 2. Since qq is between 1 and 9, inclusive, then 3+2+q3 + 2 + q is between 6 and 14. Since its ones digit is 2, then 3+2+q=123 + 2 + q = 12 and so q=7q = 7. This also means that there is a carry of 1 into the tens column. From the tens column, we see that 1+6+p+81 + 6 + p + 8 must have a ones digit of 4. Since pp is between 1 and 9, inclusive, then 1+6+p+81 + 6 + p + 8 is between 16 and 24. Since its ones digit is 4, then 1+6+p+8=241 + 6 + p + 8 = 24 and so p=9p = 9. This also means that there is a carry of 2 into the hundreds column. From the hundreds column, we see that 2+n+7+52 + n + 7 + 5 must have a ones digit of 0. Since nn is between 1 and 9, inclusive, then 2+n+7+52 + n + 7 + 5 is between 15 and 23. Since its ones digit is 0, then 2+n+7+5=202 + n + 7 + 5 = 20 and so n=6n = 6. This also means that there is a carry of 2 into the thousands column. This means that m=2m = 2. Thus, we have m+n+p+q=2+6+9+7=24m + n + p + q = 2 + 6 + 9 + 7 = 24.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.