Problem 1

Thanks to Sanket for the suggestion! This is adapted from the JEE Mains 2022 test in India.

What is the remainder when 2023 to the power 2023 is divided by 35?

Problem 2

What is the units digit of floor[10²⁰⁰⁰⁰/(10¹⁰⁰ + 3)] ?

This is from the 1986 Putnam, A2.

As usual, watch the video for solutions.

2 number theory problems from the JEE Mains and Putnam

Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Remainder 2023 To 2023 Divided By 35

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Problem 1

The letters k, N, λ, j are used to denote arbitrary integers.

Method 1: Binomial Theorem

Notice:

2023 = 2030 – 7 = 35(58) – 7 = 35k – 7

2023²⁰²³
= (35k – 7)²⁰²³
= C(2023, 0) (35k)²⁰²³(-7)⁰ + C(2023, 1) (35k)²⁰²²(-7)¹ + … + C(2023, 2023) (35k)⁰(-7)²⁰²³

All terms but the last will have a non-negative power of 35k, so all terms but the last are a multiple of 35. The equation then simplifies to:

= 35j – 7²⁰²³

So we will have the same remainder as 7²⁰²³. Now we do some more magic.

-7²⁰²³
= -7 × (7)²⁰²²
= -7 × (7²)¹⁰¹¹
= -7 × (50 – 1)¹⁰¹¹

When we expand the binomial, all terms but the very last will be multiples of 50. Multiplied by -7 will give a multiple of -350. The very last term will be (-1)¹⁰¹¹ = -1. So this simplifies to

-7 × (50 – 1)¹⁰¹¹
= -350N – 7(-1)
= 35λ + 7

This number is 7 more than a multiple of 35, and we know the original number had the same remainder upon 35.

Therefore 2023²⁰²³ has a remainder of 7 when divided by 35.

Method 2: Modulo

Since 35 = 7×5 let us look at modulo 7 and 5.

2023 = 289× 7 ≡ 0 (mod 7)
2023²⁰²³ ≡ 0 (mod 7)

2023 = 404× 5 + 3 ≡ 0 (mod 5)
2023²⁰²³ ≡ 3²⁰²³ (mod 5)

3²⁰²³ (mod 5)
≡ 3 × (3²)¹⁰¹¹ (mod 5)
≡ 3 × (-1)¹⁰¹¹ (mod 5)
≡ -3 (mod 5)
≡ 2 (mod 5)

But we also have:

2023²⁰²³ ≡ 0 (mod 7)
Thus
2023²⁰²³ = 7k ≡ 2k (mod 5)

So we have 2k ≡ 2 (mod 5), meaning k ≡ 1 (mod 5) or k = 5j + 1.

Substituting back gives:

2023²⁰²³
= 7k
= 7(5j + 1)
= 35j + 7
≡ 7 (mod 35)

Thus there is a remainder of 7.

Problem 2

(image)

Let u = 10¹⁰⁰. So we have:

10²⁰⁰⁰⁰/(10¹⁰⁰ + 3)
= u²⁰⁰/(u + 3)
= u¹⁹⁹/(1 + 3/u)

Clearly 3/u is between 0 and 1, so we can use the formula for an infinite geometric series with a common ratio -3/u.

u¹⁹⁹(1/(1 + 3/u))
= u¹⁹⁹(1 – (3/u) + (3/u)² + … + (-3/u)^k) + …)
= u¹⁹⁹ – 3u¹⁹⁸ + 3²u¹⁹⁷ + … + (-3)¹⁹⁸u + (-3)¹⁹⁹ + (-3)²⁰⁰/u + (-3)²⁰¹/u² + …
= [u¹⁹⁹ – 3u¹⁹⁸ + 3²u¹⁹⁷ + … + (-3)¹⁹⁸u] + (-3)¹⁹⁹ + [(-3)²⁰⁰/u + (-3)²⁰¹/u² + …]

The last line shows the sum bracketed in 3 groups. Let the entire number be equal to A.

The first bracketed part has positive powers of u = 10¹⁰⁰, so each term has many factors of 10, and the sum is equivalent to 0 mod 10. The last bracketed group is an alternating sum with a positive first term. So that entire sum is less than the first term.

(-3)²⁰⁰/u + (-3)²⁰¹/u² + …
< (-3)²⁰⁰/u
= (-3)²⁰⁰/10¹⁰⁰
= (9)¹⁰⁰/10¹⁰⁰
= (9/10)¹⁰⁰
< 1

So this entire bracketed sum is a positive number less than 1. This will vanish when we take the floor of the number A. So if we take the floor of A modulo 10, the only term that remains is (-3)¹⁹⁹. This is exactly the units digit of the number we want.

floor A ≡ (-3)¹⁹⁹ (mod 10)

But now this calculation is straight-forward.

(-3)¹⁹⁹ (mod 10)
≡ (-3)^{4×49 + 3} (mod 10)
≡ (-3)^4×49(-3)³ (mod 10)
≡ ((-3)⁴)⁴⁹(-3)³ (mod 10)
≡ (1)⁴⁹(3) (mod 10)
≡ 3 (mod 10)

Thus the units digit is 3.

References

Problem 1
Brainly
https://brainly.in/question/55364464
Sarthaks
https://www.sarthaks.com/3443400/the-remainder-when-2023-2023-is-divided-by-35-is
StackExchange
https://math.stackexchange.com/questions/4668677/remainder-when-20232023-is-divided-by-35

Problem 2
John Scholes Putnam problems
https://prase.cz/kalva/putnam/psoln/psol862.html
StackExchange
https://math.stackexchange.com/questions/294400/rightmost-digit-of-left-lfloor-frac1020000101003-right-rfloor
YouTube vids
https://youtu.be/3eT5OKPOVvE?si=phkyzREdiHX3AbMk
https://www.youtube.com/watch?v=Phf8QyAXop8
https://www.youtube.com/watch?v=m7Wq33Q9hJ8
https://www.youtube.com/watch?v=3eT5OKPOVvE

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