CSAT Solved Papers/ 2023/Q49

2023 CSAT — Q49

Quant Logical & quantitative reasoning 2.5 marks Hard

If today is Sunday, then which day is it exactly on 101010^{10} th day?

  1. A Wednesday Answer
  2. B Thursday
  3. C Friday
  4. D Saturday

Worked rationale

Number the days with today (Sunday) as the 11st day. Then the 101010^{10}th day is 1010−110^{10} - 1 days after today, so the weekday is Sunday shifted by (1010−1) mod 7(10^{10} - 1) \bmod 7.

Compute 1010 mod 710^{10} \bmod 7. Since 10≡3(mod7)10 \equiv 3 \pmod 7, use the cycle of 33:

31≡3, 32≡2, 33≡6, 34≡4, 35≡5, 36≡1(mod7).3^1\equiv 3,\ 3^2\equiv 2,\ 3^3\equiv 6,\ 3^4\equiv 4,\ 3^5\equiv 5,\ 3^6\equiv 1 \pmod 7.

The order is 66; 10≡4(mod6)10 \equiv 4 \pmod 6, so 1010≡310≡34≡4(mod7)10^{10} \equiv 3^{10} \equiv 3^{4} \equiv 4 \pmod 7.

Thus (1010−1)≡4−1=3(mod7)(10^{10} - 1) \equiv 4 - 1 = 3 \pmod 7. Sunday +3+ 3 days == Wednesday.

Sun→+1Mon→+2Tue→+3Wed.\text{Sun} \xrightarrow{+1}\text{Mon}\xrightarrow{+2}\text{Tue}\xrightarrow{+3}\text{Wed}.

Answer: (a) Wednesday.

Why the other options miss

  • B
    a convention slip: counts 101010^{10} days after today (offset 1010≡410^{10}\equiv 4, giving Sun+4=+4 = Thursday) instead of treating the 101010^{10}th day as position 101010^{10} (offset 1010−110^{10}-1).
  • C
    an arithmetic slip: reduces the exponent wrongly (e.g. 10 mod 6=510 \bmod 6 = 5), landing on 35≡53^5 \equiv 5, Sun+(5−1)=+ (5-1) = Friday-ish.
  • D
    misapplies the cycle-length rule: uses the wrong repeating cycle for powers of 1010 mod 77 (mistaking the order 66 for 77, or reducing the exponent against the wrong modulus), drifting the offset by a day.

Specialist insight

Two disciplines: (1) reduce the power mod 77 via the order of 1010, which is 66 — so reduce the exponent 1010 mod 66 to get 1010≡34≡410^{10}\equiv 3^4 \equiv 4; (2) read the phrase precisely — the “101010^{10}th day” is the day at position 101010^{10} counting today as the 11st, an offset of 1010−1≡310^{10}-1\equiv 3. The Thursday distractor is the ”NN days later” reading; the scoring move is matching the offset convention to the wording, landing on Wednesday.

The trap, in one line

ord7(10)=6\mathrm{ord}_7(10)=6, 1010≡34≡410^{10}\equiv 3^4\equiv 4; the 101010^{10}th day is offset 4−1=34-1=3 from Sunday ⇒\Rightarrow Wednesday (a).

← All 2023 CSAT questions