Number Theory

By the second digit from the left, I assume you mean 5n451 where n is the second digit from the left

1

3

5

7

9

5 ways for the second digit from the left

1 (0 or 1 for the last digit)

3 (0 or 1 for the last digit)

5 (0 or 1 for the last digit)

7 (0 or 1 for the last digit)

9 (0 or 1 for the last digit)

10 ways for the second digit and last digit

Now the first digit can be 1 through 9 so 10*9 => 90

So the first digit, the second digit from the left, and the last digit is taken care of.

The 3rd and 4th digit can be 100 ways as it can be 00 to 99.

So 90*100 => 9000???

I dont know I came up this quickly I dont have much time.

I wrote a relatively short computer code just to see what numbers would I get:

n=0;p=0;cycle:a=(10000+n);b=int(a/10000);c=int(a/1000);d=c%10;e=int(a/100);f=e%10;g=int(a/10);h=g%10;i=int(a/10);j=a%10;n=n+5;if(b!=d and b!=f and b!=h and b!=j and d!=f and d!=h  and d!=j and f!=h and f!=j and h!=j, goto loop,goto cycle);loop:if(d%2==0, goto cycle,0);p=p+1;printa," ",;if(n<87870, goto cycle, 0);print"Total = ",p

It begins as follows: 13025  13045  13065  13075  13085  13095  13205  13240  13245  13250  13260  13265  13270  13275  13280  13285  13290  13295  13405  13420  13425  13450  13460  13465  13470  13475  13480  13485............... And ends as follows.............97610  97615  97620  97625  97630  97635  97640  97645  97650  97680  97685  97805  97810  97815  97820  97825  97830  97835  97840  97845  97850  97860  97865  Total =  2856

Note: See if anybody can verify the above total.

There are = 1,680 numbers ending in "0"
There are = 1,176 numbers ending in "5"
There are 294 beginning with "1", of which 168 end in "0" and 126 end in "5". This pattern holds true for all odd numbers, except for 5 which has only 168 numbers ending in "0"
There are 378 beginning with "2" of which 210 end in "0" and 168 end in "5". Again this holds true for all even numbers.
There are 168 beginning with "5" and ALL ending in "0"
There are 126 x 4  +  168 x 4 =1,176 that end in "5"
There are 168 x 5   +  210 x 4=1,680 that end in "0"
126 consists of the following occurences:1 x 1 x 3 x 6 x 7=126
168 =1 x 1 x 4 x 6 x 7 = 168
210 =1 x 1 x 5 x 6 x 7 =210
Therefore: 1,176 + 1,680 =2,856 numbers is accurate.

How many 5-digit numbers are there which are multiples of 5 and are ALL different and that the 2nd digit from the left is odd? Thanks for any help.

9 choices, 5 choices, 10 choices, 10 choices, 2 choices

9*5*10*10*2 = 9000

What are all different?  All these 9000 numbers are different ......

Hi Melody: Your count includes many duplicate numbers such as:1 1 4 5 0. Notice the duplicate "1". This number is allowed only if it is: 1 3 4 5 0. All 5 numbers must be different and the 2nd digit from the left must be odd. You start with 9 digits from the left. But one of them, 3, cannot be included because you will have 2 "3s", which is not allowed. That leaves 8 digits. But the first digit cannot be "4" or "5" because you will have duplicates. That leaves only 6 numbers: 1, 2, 6, 7, 8, 9. And so on. In the above computer code, I have included all these exceptions in and have a full list of all allowable numbers. They total =2,856 such numbers.