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Determine the integer n with the properties :
1.)n is a prime less than 6000
2.)The number formed by the last two digits of n is < 10
3.)If the decimal digits of n are reversed to obtain N , then N - n = 999
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2) means that the second last digit must be 0, so you have a number of the form
ab0c (not multplication) = 1000a+100b+c (a,b,c are integers in the range 0-9)
3) the reversal N is c0ba = 1000c+10b+a
N-n = 1000(c-a)-100b+10b+(a-c) = 999
a-c = 9
b = 9
hmmm. im a bit lost with it now, the negatives are messing things up in my head here
Last edited by luca-deltodesco (2007-11-16 22:23:23)
The Beginning Of All Things To End.
The End Of All Things To Come.
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Hence N is the number obtained from n by adding 1 to the first digit and subtracting 1 from the last digit of n. (NB: n doesnt end with two 0s otherwise it wouldnt be prime.)
∴ c = a+1, b = 0.
Also, c (being the last digit of a 4-digit prime) must be 1, 3, 7 or 9. But c cant be 1, 7 or 9 because the first digit of n is c−1 and n is < 6000. This leaves only c = 3.
Hence:
Last edited by JaneFairfax (2007-11-17 02:20:37)
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Practical arithmetic tip:
Adding 999 to a 4-digit number whose last digit is not 0 is the same as adding 1 to the first digit and subtracting 1 from the last digit.
Similarly for adding 99, 9999, 99999, etc. Thats how I add up my shopping cost to find out how much to pay before going to the cashiers (just to make sure I dont overspend). If something costs 99p and I add it to my shopping basket, I mentally add £1 and deduct 1p from my shopping total.
Example: £5.34 + 99p = £5.34 + £1 − 1p = £(5+1).(34−1) = £6.33
And this can be modified to use on more complicated sums:
£5.34 + £2.93 = £5.34 + £3 − 7p = £(5+3).(34−7) = £8.27
Last edited by JaneFairfax (2007-11-17 01:09:13)
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