1234 Followers. A problem for "1234"

If the total number of 8 8 -digit positive integers that can be formed using the digits 0 0 to 9 9 inclusive and without repetition. such that 1234 \overline{1234} always exist collectively in the same order in the positive integer is P P .

We coin the term: Cute divisors . A cute divisor of a number n n is a positive integer other than 1 1 and n n which divide the number n n completely and leaves no remainder.

Find the sum of all cute divisors of P P .


The answer is 3479.

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2 solutions

Pawan Kumar
Apr 18, 2015

Since 1234 1234 must exist contiguously, we can consider them as a single digit A A .

Since the original number is 8 8 digit, the number now reduces to 5 5 digit numbers ( A 4 A 3 A 2 A 1 A 0 ) (A_4A_3A_2A_1A_0) that can be formed using digits { 0 , A , 5 , 6 , 7 , 8 , 9 } \{0, A, 5, 6, 7, 8, 9\} .

Total ways to place these 7 7 digits in a 5 5 digit number, N ( T ) N(T) = 7 × 6 × 5 × 4 × 3 = 2520 =7 \times 6 \times 5 \times 4 \times 3 = 2520

Total 5 5 digit numbers that had A 4 A_4 as 0 0 , N ( Z ) N(Z) = 1 × 6 × 5 × 4 × 3 = 360 = 1 \times 6 \times 5 \times 4 \times 3 = 360

Total 5 5 digit numbers that did not have A A , N ( A ) N(\overline{A}) = 6 × 5 × 4 × 3 × 2 = 720 = 6 \times 5 \times 4 \times 3 \times 2 = 720

Total 5 5 digit numbers that did not have A A and had A 4 A_4 as 0 0 , N ( A Z ) N(\overline{A} \cap Z) = 1 × 5 × 4 × 3 × 2 = 120 = 1 \times 5 \times 4 \times 3 \times 2 = 120

\therefore Total valid 5 5 digit numbers that have A A , P P

= N ( T ) N ( Z ) N ( A ) + N ( A Z ) = N(T) - N(Z) - N(\overline{A}) + N(\overline{A} \cap Z)

= 2520 360 720 + 120 = 2520 - 360 - 720 + 120

= 1560 = 1560

Prime factorization of P = 2 3 × 3 × 5 × 13 P = 2^3 \times 3 \times 5 \times 13

Sum of divisors of P = 2 4 1 2 1 × ( 3 + 1 ) × ( 5 + 1 ) × ( 13 + 1 ) = 5040 P = \frac{2^4-1}{2-1} \times (3 + 1) \times (5+1) \times (13 + 1) = 5040

Sum of cute divisors of P = 5040 1 1560 = 3479 P = 5040 - 1 - 1560 = 3479

Vishnu C
Mar 24, 2015

Hey Sandeep! I checked on the internet and it says that 1 is also considered a proper divisor. I thought you meant the sum of proper divisors excluding one and I subtracted 1 when I entered 3480 and it said that it was wrong. The solution is simple enough: 1234 can be considered a block occupying one space and the other digits occupying one space each. This reduces it to a sort of 5 digit problem. 0 should not occupy the first place and we don't have to worry about this when the 1234 block occupies the 1st place. We then get: (5!×5×4+6!)/2=1560=8×3×5×13. Sum of divisors of 1560=5040. Sum of proper divisors (excluding 1) = 3479.

I couldn't understand your step in which you got P .Can you elaborate,?

A Former Brilliant Member - 6 years, 2 months ago

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Since 1234 always remain together, they can be considered as a single unit. Hence the number may look like any of the following (i) 1234 _ _ _ _ (ii) _ 1234 _ _ _ (iii) _ _ 1234 _ _ (iv) _ _ _ 1234 _ (v) _ _ _ _ 1234
In the first choice we have (6×5×4×3) cases, in the rest of the choices, 0 can't be in the first position, hence number of cases in each choice (5×5×4×3) , so total number of cases (360 + 4*300=1560)

Agniprobho Mazumder - 6 years, 2 months ago

yes, 1 is a proper divisor so, i think answer should be 3480. @Sandeep Bhardwaj

Karan Siwach - 6 years, 2 months ago

But the question says "cute divisors" not "proper divisors" .

Anik Mandal - 6 years, 2 months ago

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question has been edited.

Karan Siwach - 6 years, 2 months ago

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