The binomial coefficient 99!/(80!*19!) is equal to the 21-digit number: 107,196,674,080,761,936,ABC.
Using pen and paper- find the three missing digits A, B and C.
The number can be expressed as a fraction whose numerator is the product of the numbers 81-99 and the denominator is the product of the numbers 1-19.
The numerator has two multiples of 11: 99 and 88. The denominator only one multiple of 11: 11. Then the number is a multiple of 11.
The numerator has seven multiples of 3, three of which are multiples of 9: 99, 96, 93, 90, 87, 84, 81. The denominator has six multiples of 3, two of which are multiples of 9: 18, 15, 12, 9, 6, 3. Then the number is a multiple of 9.
The numerator has three multiples of 7, one of which is a multiple of 49: 98, 91, 84. The denominator has two multiples of 7: 14 and 7. Then the number is a multiple of 49 and 7.
The number can be expressed as 107,196,674,080,761,936,000 + ABC.
107,196,674,080,761,936,000 is a multiple of 11: 1-0+7-1+9-6+6-7+4-0+8-0+7-6+1-9+3-6+0-0+0 = 11. Then ABC must be a multiple of 11.
107,196,674,080,761,936,000 is a multiple of 9: 1+0+7+1+9+6+6+7+4+0+8+0+7+6+1+9+3+6+0+0+0 = 81 = 9*9. Then ABC must be a multiple of 9.
107,196,674,080,761,936,000 is congruent to 1 mod 7: 107-196+674-080+761-936+000 = 330 = 47*7+1. Then ABC must be congruent to 6 mod 7.
594 is a multiple of 9 and 11 and is congruent to 6 mod 7. Therefore ABC = 594.
Solution
