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How many times has 126299250 been multiplied, and the last five math words are all 0?
Remember x=ABCDE, y=(*)(*)(*)(*)(*) (*) (*).

Then x * x =100000 * y+X.

x*x-x = 100000*y

x*(x- 1)= 100000*y

y=(x*x-x)/ 100000=x*(x- 1)/(2^5*5^5)

X is a multiple of 5 = 3 125 (because X is a five-digit number, it cannot be a multiple of 10 5 = 100000).

Let x=3 125k (k is a natural number).

Then 3125k * (3125k-1) =100000 * y.

(3 125k- 1) is a multiple of 2 5 = 32.

Let (3 125k- 1) =32m (m is a natural number).

m =(3 125k- 1)/32 = 98k-( 1 1k+ 1)/32

Let (11k+1)/32 = n (n is a natural number).

rule

k =(32n- 1)/ 1 1 = 3n-(n+ 1)/ 1 1

N =-1+11c (c is a natural number)

n=- 1 k=-3

n= 10 k=30- 1=29,x=90625

n=2 1 k=63-2=6 1,x = 190625 & gt; 100000 give up

N=32 k=96-3=93, x = 290625 & gt 100000 give up.

So x=90625

A=9,B=0,C=6,D=2,E=5