Problem 37: The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
Continued reading >
Problem 36: Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.
Continued reading >
Problem 35: How many circular primes are there below one million?
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Problem 34: Note: 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Since 1! = 1 and 2! = 2 are not sums they are not included.
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Saturday, August 4th, 2012
Problem 33:
The fraction 49/98 is a curious fraction, as 49/98 = 4/8, obtained by cancelling the 9s.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
Continued reading >
Saturday, August 4th, 2012
Problem 32: Find the sum of all numbers that can be written as pandigital products.
Continued reading >
Problem 31: Given the following coin values: (1p,2p,5p,10p,20p,50p,£1,£2):
Where (1p = 1/£1)
How many different ways can £2 be made using any number of coins?
Continued reading >
Thursday, August 2nd, 2012
Problem 30: Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
Continued reading >
Thursday, August 2nd, 2012
Problem 29: How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
Continued reading >
Thursday, August 2nd, 2012
Problem 28: Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
Continued reading >