+0

-2
191
6
+397

deleted.

Oct 29, 2019
edited by sinclairdragon428  Nov 20, 2019

#1
+1

There are: 5^4 = 625 odd integers

There are 9000 / 5 =1,800 multiple of 5 integers

900 + 625 =1525 Total integers

Oct 29, 2019
#2
+24364
+3

let a equal the number of four digit odd numbers. let b equal the number of four digit multiples of 5. find a+b.

I assume:

These are two arithmetic series.

$$\begin{array}{|rcll|} \hline 1001 + (a-1)\cdot 2 &=& 9999 \quad | \quad \small \text{first odd four digit number }=1001,\ \text{last odd four digit number }=9999\\ (a-1)\cdot 2 &=& 9999 - 1001 \\ (a-1)\cdot 2 &=& 8998 \\ a-1 &=& 4499 \\ a &=& 4499+1 \\ \mathbf{a} &=& \mathbf{4500} \\\\ 1000 + (b-1)\cdot 5 &=& 9995 \quad | \quad \small \text{first four digit number multiples of 5}=1000,\ \text{last four digit number }=9995\\ (b-1)\cdot 5 &=& 9995 - 1000 \\ (b-1)\cdot 5 &=& 8995 \\ b-1 &=& 1799 \\ b &=& 1799+1 \\ \mathbf{b} &=& \mathbf{1800} \\ \hline \end{array}$$

$$a+b = 4500+1800 = 6300$$

Oct 29, 2019
#3
0

heureka: Isn't the number of odd integers from the left: 5(1,3,5,7,9) x 5 x 5 x 5 =5^4 =625 such numbers.

Oct 29, 2019
#4
0

heureka: You counted 4,500 ODD digits. He/she wants "four-digit odd numbers", which is 5^4 =625 odd numbers.

Guest Oct 29, 2019
#5
+397
+1

Thank you to everyone! 6300 is the answer, so special thanks to heureka

Oct 29, 2019
#6
+24364
+2

Thank you, sinclairdragon428 !

heureka  Oct 29, 2019