Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:
D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins
21
A takes 6 hours to fill
divide 100 by 7, and you get 14.28. (Obviously you aren’t talking about decimals here) and so 14 numbers can be divisible by 7 up to 100.
The answer is 14.
FYI
Answer:
24
Step-by-step explanation:
A + B = 40.
And at Rs 7 a kg for 40 kg, you want a total of Rs.280.
So the second equation is 9A + 4B = 280
From the first equation: B = 40 – A
and sub into the second equation:
9A + 4(40-A) = 280
9A + 160 – 4A = 280
5A = 120
A = 24.
And you should check: B should equal 40-24 = 16. Check with the final equation: 9*24 + 4*16 = 216 + 64 = 280. So it works.
Your answer, of course, is A = 24
First we need to find out LCM of 2,3,5
that is 30,,,
then add 30 to 6 we get 36…
then divide it by 2 we get 18..
so 18 would be written interms of binay as 10010
means..Answer is
$**$*
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
12