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# Sum of Digits base 10 | PRMO II 2019 | Question 7

Try this beautiful problem from the PRMO II, 2019 based on the Sum of Digits base 10. You may use sequential hints to solve the problem.

Try this beautiful problem from the PRMO II, 2019 based on Sum of Digits base 10.

## Sum of Digits base 10 – PRMO II 2019

Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m)=20 and s(33m)=120, what is the value of s(3m)?

• is 107
• is 60
• is 840
• cannot be determined from the given information

### Key Concepts

Real Numbers

Algebra

Integers

But try the problem first…

Source

PRMO II, 2019, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

taking sum of digit base 10 to (mod 9)

and s(ab)=s(a).s(b)(mod 9)

[ let x congruent r mod n, y congruent to s mod n,

$$0 \leq r,s \leq n-1$$,

x=in+r, y=jn+s, i,j are integers

xy=(in+r)(jn+s)=ij$$n^2$$+(is+jr)n+rs congruent to rs mod n

so, xy mod n =(x mod n)(y mod n) ]

Second Hint

given s(m)=20

s(33m)=120=$$s(11) \times s(3m)$$

or, 120=$$2 \times s(3m)$$ [ since s(11)=2(mod 9)]

Final Step

or, 60=s(3m)

so, s(3m)=60.

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