More questions?

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more questions???

Hi more questions again... :blushing: :blushing: i am having a hard time figuring out these questions...
1)a)how many 6-digit palindromic numbers are there???
(i can do this the slow way, but i am looking for shortcuts)
b) how many odd 7-difit palindromic numbers are there in which every digit appears at most twice?
( again, i am looking for a fast way)

2)show that if 14 distinct intergers are chosen form the sequence 100, 101,102, 103......,123,124, there must be two of them whose difference is 4
(i have worked this out by writing out 14 numbers and having all the possibilities(it took a long time) and i am looking for a shortcut) :smile:
 

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HallsofIvy
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chickenguy said:
Hi more questions again... :blushing: :blushing: i am having a hard time figuring out these questions...
1)a)how many 6-digit palindromic numbers are there???
(i can do this the slow way, but i am looking for shortcuts)
What do you mean by the "slow way". Since the numbers are palindromic, once you have the first 3, the others must be the same. There are 103 such numbers.

b) how many odd 7-difit palindromic numbers are there in which every digit appears at most twice?
( again, i am looking for a fast way)
Lazy , eh? Since 7 is odd, we are looking for 4 digits. No digit more than twice? Okay, there are 10 possible first digits, then 10 possible second digits (which might be the same). We might want to separate those: there are 10(9)= 90 two different digit combos, 10 possible where the two digits are the same. Of the "different" two digits, there are again 10 possiblilities for the 3 digit but again we will want to consider separately that digit being the same or not the same as one of the first two... If the first two digits are the same then there are 9 possibilities for the third digit, etc.
I don't see any "easy" way. Sorry.

2)show that if 14 distinct intergers are chosen form the sequence 100, 101,102, 103......,123,124, there must be two of them whose difference is 4
(i have worked this out by writing out 14 numbers and having all the possibilities(it took a long time) and i am looking for a shortcut) :smile:
pigeon hole principal ought to work. How many different "differences" are possible?
 
Alkatran
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chickenguy said:
2)show that if 14 distinct intergers are chosen form the sequence 100, 101,102, 103......,123,124, there must be two of them whose difference is 4
(i have worked this out by writing out 14 numbers and having all the possibilities(it took a long time) and i am looking for a shortcut) :smile:
Hint: There's a reason the range is from N to N+24 (includes less than 14*2 - 1 integers)
 
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HallsofIvy said:
What do you mean by the "slow way". Since the numbers are palindromic, once you have the first 3, the others must be the same. There are 103 such numbers.
Actually, wouldn't it be 900, since the first digit can't be 0?
 

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