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Ques:- A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager. According to his will: First give one coin to manager. 1/5th of the remaining to the elder son.Now give one coin to the manager and 1/5th of the remaining to second son and so on….. After giving coins to 5th son, divided the remaining coins among five daughters equally.All should get full coins. Find the minimum number of coins he has?
Recent Answer : Added by Biplab Nath On 2021-07-31 16:35:53:
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
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Recent Answer : Added by Admin On 2020-05-17 12:01:52:
The answer probably lies in finding the in-centre of the
traingle. Bisect all the angles of the triangles and the
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Recent Answer : Added by Bipul On 2022-03-09 16:42:16:
130
520 = 26 * 20 = 2 * 13 * 22 * 5 = 23 * 13 * 5
Required smallest number = 2 * 13 * 5 = 130
130 is the smallest number which should be multiplied with 520 to make it a perfect square.
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Recent Answer : Added by Hamiss On 2022-09-29 16:38:02:
Palwal
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Recent Answer : Added by Amudha A On 2021-09-09 07:24:42:
A=P+SI——(1)
sum of money after 30 years = Double the money
A=2P———-(2)
Equate 1 and 2
P+SI=2P
SI=2P-P
SI=P————-(3)
WKT, SI= PTR/100
Equate 3 and SI
P=PTR/100
P=P*30*R/100
R=P*100/P*30
R=100/30 or 10/3 or 3(1/3) %
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Recent Answer : Added by Admin On 2022-09-28 17:11:47:
No of men employed = N
No of days to finish the work = 9 days
No of men after increase = (N + 10)
No of days to finish the work = 6 days
Equating mandays
9N = (N+10)*6
9N — 6N = 60
3N = 60
N = 20
No of men employed = 20
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Recent Answer : Added by Akshay On 2022-03-09 16:46:00:
225
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5A=3B,
*3 SO,
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Hog
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Recent Answer : Added by Deepikaa Nandakumar On 2022-05-17 09:08:52:
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Recent Answer : Added by nani Reddy On 2021-08-05 17:11:29:
avg sal of 3 is $950
first person earns=$1150
second person earns=$650
let thirdperson =x
avg=(1150+650+x)/3
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x=2850-1800
x=$1050
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Recent Answer : Added by Admin On 2020-05-17 12:02:21:
((15 X m) + 23)/(m+1)= 16
solving , we get number of matches (m) as 7 (excluding last
match)
so he shud hit 39 runs in last innings to ake average to 18
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