N = aebfcg is a six digit number where a, b, c, e, f, g are its six digits. If $a = e$, $b = f$ and $c = g$, which of the given statements is NOT correct?

Question

Let the number $(22)^{2022}$ + $(2022)^{22}$ leave the remainder $ \alpha $ when divided by 3 and $ \beta $ when divided by 7. Then $ (\alpha^2 + \beta^2) $ is equal to:}

Answer 1

Instead of proving things in general, let's test this by trying real numbers and see which statement breaks down.

Since the digits of N are really a, a, b, b, c, c (because e = a, f = b, g = c), we can always write $N = 110000a + 1100b + 11c$, which is $11$ times $(10000a + 100b + c)$. This means N is a multiple of $11$ no matter what a, b, c are.

Option 1: Take g = c = 4, and try a = 1, b = 0. Then N = $110044$, and $110044 / 44 = 2501$ exactly, no remainder. Try a = 2, b = 3: N = $223344$, and $223344 / 44 = 5076$, also exact. The pattern always holds because $100$ and $10000$ are already multiples of $4$, so having $c = 4$ is all it takes. This statement is correct.
Option 2: Take a = 1, b = 2, c = 3, so a + b + c = 6. Then N = $112233$, and $112233 / 33 = 3401$ exactly. Try a = 2, b = 2, c = 2 (sum 6 again): N = $222222$, and $222222 / 33 = 6734$, exact again. This statement is correct.
Option 3: Take g = c = 8, and try several pairs (a, b). With a = 1, b = 0, N = $110088$; the closest squares are $331^2 = 109561$ and $332^2 = 110224$, so $110088$ is not a square. With a = 5, b = 5, N = $550588$; the closest squares are $741^2 = 549081$ and $742^2 = 550564$, still not a square. No matter which a and b are tried, N always ends in the digits 0 and 8, and a perfect square that is 11 times another number can never end in those two digits, once you check the pattern of last two digits that squares can take. So N can never be a perfect square when g = 8, making this statement FALSE.
Option 4: Take b = c = 0, so N = $110000a$ for a = 1 to 9, giving $110000, 220000, \ldots, 990000$. None of these are perfect squares: for example $\sqrt{110000} \approx 331.7$ and $\sqrt{990000} \approx 995.0$, neither landing on a whole number, and the same holds for every a from 1 to 9. This statement is correct.

Only option 3 breaks down under testing, since g = 8 never allows N to become a perfect square for any choice of a, b, c.

Let's summarize:

N always has the form $11 \times (10000a + 100b + c)$, so it is always a multiple of 11.
Options 1, 2 and 4 hold up under direct testing with real numbers.
Option 3 fails: no choice of a, b, c ever makes N a perfect square when g = 8.

So the statement that is NOT correct is option 3.

Answer 2

Instead of proving things in general, let's test this by trying real numbers and see which statement breaks down.

Since the digits of N are really a, a, b, b, c, c (because e = a, f = b, g = c), we can always write $N = 110000a + 1100b + 11c$, which is $11$ times $(10000a + 100b + c)$. This means N is a multiple of $11$ no matter what a, b, c are.

Option 1: Take g = c = 4, and try a = 1, b = 0. Then N = $110044$, and $110044 / 44 = 2501$ exactly, no remainder. Try a = 2, b = 3: N = $223344$, and $223344 / 44 = 5076$, also exact. The pattern always holds because $100$ and $10000$ are already multiples of $4$, so having $c = 4$ is all it takes. This statement is correct.
Option 2: Take a = 1, b = 2, c = 3, so a + b + c = 6. Then N = $112233$, and $112233 / 33 = 3401$ exactly. Try a = 2, b = 2, c = 2 (sum 6 again): N = $222222$, and $222222 / 33 = 6734$, exact again. This statement is correct.
Option 3: Take g = c = 8, and try several pairs (a, b). With a = 1, b = 0, N = $110088$; the closest squares are $331^2 = 109561$ and $332^2 = 110224$, so $110088$ is not a square. With a = 5, b = 5, N = $550588$; the closest squares are $741^2 = 549081$ and $742^2 = 550564$, still not a square. No matter which a and b are tried, N always ends in the digits 0 and 8, and a perfect square that is 11 times another number can never end in those two digits, once you check the pattern of last two digits that squares can take. So N can never be a perfect square when g = 8, making this statement FALSE.
Option 4: Take b = c = 0, so N = $110000a$ for a = 1 to 9, giving $110000, 220000, \ldots, 990000$. None of these are perfect squares: for example $\sqrt{110000} \approx 331.7$ and $\sqrt{990000} \approx 995.0$, neither landing on a whole number, and the same holds for every a from 1 to 9. This statement is correct.

Only option 3 breaks down under testing, since g = 8 never allows N to become a perfect square for any choice of a, b, c.

Let's summarize:

N always has the form $11 \times (10000a + 100b + c)$, so it is always a multiple of 11.
Options 1, 2 and 4 hold up under direct testing with real numbers.
Option 3 fails: no choice of a, b, c ever makes N a perfect square when g = 8.

So the statement that is NOT correct is option 3.

Over Number	Run Rate
N-2	8.00
N	7.43
N+2	8.11
N+4	8.45
N+6	8.08

N = aebfcg is a six digit number where a, b, c, e, f, g are its six digits. If \(a = e\), \(b = f\) and \(c = g\), which of the given statements is NOT correct?

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Divisibility and Remainder

Questions Asked in XAT exam