A number theory problem by Anthony Pham
There is a special six-digit number n such that:
-its first two digits( a ), its middle two digits( b ), and its last two digits( c ) each form a two-digit prime number
- a < b < c
- c is greater than a by 60
-the units digits of b is 3
-it has three two-digit prime factors, which if written out from least to greatest, it is greater than n by 100000
What is the six-digit number n ?
The answer is 378397.
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1 solution
Since c is a two-digit number and is greater than a by 60, a must be at most 37, and c must be at least 71, since a is two digits, so a is also at least 11. As well, a and c will have the same units digit, so a is of the set:{11, 13, 19, 23, 29, 37}, and c is of the set:{71, 73, 79, 83, 89, 97}. ( a is not equal to 17 nor 31 since c would be 77 and 91, respectively). You can also find the set:{13, 23, 43, 53, 73, 83} for b .
Since the three two-digit prime factors written out would be greater than n by 100000, b and c are two of its prime factors. a + 10 is yet another factor. This means that the units digit of abc is equal to the units digit of n . We know that the units digit of b is 3, and that the units digit of c must be consistent. Since a has the same units digit as c , then we try it for 1, 3, 7, and 9.
(1 mod 10)²(3 mod 10)=3 mod 10
(3 mod 10)²(3 mod 10)=7 mod 10
(7 mod 10)²(3 mod 10)=7 mod 10
(9 mod 10)²(3 mod 10)=3 mod 10
Therefore, a and c have 7 as its units digit; a is 37, and c is 97. We can then find that
47 x b x 97 = 370 b 97 (in base form)
47 x 13 x 97 = 59267
47 x 23 x 97 = 104857
47 x 43 x 97 = 196037
47 x 53 x 97 = 241627
47 x 73 x 97 = 332807
47 x 83 x 97 = 378397
n = 378397