A-7-19-31-43-upto-29-terms-find-the-last-four-digits-of-A-

Question Number 220579 by Rojarani last updated on 16/May/25

$$\:{A}=\mathrm{7}×\mathrm{19}×\mathrm{31}×\mathrm{43}×…..{upto}\:\mathrm{29}\:{terms} \\ $$$$\:{find}\:{the}\:{last}\:{four}\:{digits}\:{of}\:{A}. \\ $$

Answered by cadmon98 last updated on 16/May/25

$$ \\ $$We are given:

A = 7 × 19 × 31 × 43 × … (29 terms in total)

Each number seems to follow an arithmetic progression:

First term (a) = 7

Common difference (d) = 12
(since 19 – 7 = 12, 31 – 19 = 12, …)

So the terms are:
7, 19, 31, …, where the n-th term is:

Tₙ = 7 + (n – 1) × 12

We need to find the last 4 digits of the product of these 29 terms, i.e., A mod 10000.

Now, 10000 = 2⁴ × 5⁴ = 16 × 625

So we’ll calculate:

A mod 16

A mod 625
Then combine them using Chinese Remainder Theorem (CRT).

—

Step 1: A mod 16

We're multiplying 29 odd numbers, all of which are congruent to 7 mod 12, increasing by 12 each time.

Let's reduce each term modulo 16:

We can compute the sequence mod 16:

7 mod 16 = 7

19 mod 16 = 3

31 mod 16 = 15

43 mod 16 = 11

55 mod 16 = 7

67 mod 16 = 3

…

You can see the cycle: 7, 3, 15, 11 repeats every 4 terms.

Cycle = [7, 3, 15, 11] → product mod 16:

7×3 = 21 → 5
5×15 = 75 → 11
11×11 = 121 → 9

So one full cycle product mod 16 = 9

Each cycle = 4 terms → 29 terms = 7 full cycles (28 terms) + 1 extra term

1 extra term = next in the cycle → first term = 7

So:

7 cycles: 9^7 mod 16

Extra term: × 7

Let’s compute:

9^1 = 9

9^2 = 81 → 1 mod 16

9^3 = 9 (since powers cycle every 2)

So 9^7 mod 16 = 9
Then A mod 16 = 9 × 7 = 63 mod 16 = 15

—

Step 2: A mod 625

We want the product of 29 numbers mod 625.
But 625 = 5⁴, so if the product contains at least four 5s, the result mod 625 = 0.

Check if any of the 29 terms is divisible by 5:

Yes:

55 (5×11)

115

175

235

295

These are terms of the form:
Tₙ = 7 + 12(n − 1) ≡ 0 mod 5
→ 12(n−1) ≡ -7 mod 5 → 2(n−1) ≡ 3 mod 5 → n−1 ≡ 4 mod 5 → n ≡ 5 mod 5

So every 5th term → total = floor(29 / 5) = 5 terms divisible by 5

So the product has at least 5 factors of 5, meaning it's divisible by 5⁵ = 3125, so definitely divisible by 625

So A mod 625 = 0

—

Step 3: Use CRT

We found:

A ≡ 15 mod 16

A ≡ 0 mod 625

We want a number x ≡ 0 mod 625 and x ≡ 15 mod 16

Let x = 625k

Then: 625k ≡ 15 mod 16
→ k ≡ (15 × 625⁻¹) mod 16

Now, find the modular inverse of 625 mod 16:

625 mod 16 = 625 – 16×39 = 625 – 624 = 1
So 625 ≡ 1 mod 16 → inverse is also 1

So: k ≡ 15 × 1 ≡ 15 mod 16
→ k = 16m + 15
→ x = 625k = 625(16m + 15) = 625×15 + 625×16m
→ x = 9375 + 10000m

So A ≡ 9375 mod 10000

—

Final Answer:

The last 4 digits of A are: 9375

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