PUZZLE 10: 22nd-28th February (to be e-mailed by the 1st March)
POSTED BY IES CAMPANAR
THE MAGIC NUMBER
'Free me, please', Ali begged the genius who had trapped him in a cage.‘I will free you only if you find a number which obeys certain conditions’, the genius answered.And these were the conditions:1.- If the number were multiple of 2, then it would be a number between 50 and 59, both included.2.- If it were not a multiple of 3, then it would be a number between 60 and 69, both included.3.- If it were not a multiple of 4, then it would be a number between 70 and 79, both included.Which was the magic number?
SOLUTION
If the number is a multiple of 2, it could be:50, 52, 54, 56, 58If it is not a multiple of 3, the number could be:61, 62, 64, 65, 67, 68If it is not a multiple of 4, the possibilities are:70, 71, 73, 74, 75, 77, 78, 79By eliminating the only possibility is number 75.
Monday, February 22, 2010
Monday, February 8, 2010
FORTNIGHT'S PUZZLE 9
PUZZLE 9: 8th-14th February (to be e-mailed by the 15th February)
POSTED BY IES CAMPANAR
DIGIT NUMBERS
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.
SOLUTION
The number in polynomial form is 10a+b. Because of the condition:(10a+b)/(a+b) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
POSTED BY IES CAMPANAR
DIGIT NUMBERS
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.
SOLUTION
The number in polynomial form is 10a+b. Because of the condition:(10a+b)/(a+b) = 710a+b = 7a + 7b3a = 6ba = 2bWith b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84
Monday, February 1, 2010
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