Ek purse m 100rs and 50 rs k note h jinka total price 25000 hai if 100 k notes 50 k notes se ek Jada h to dono type k notes ki number find kro
Answers
As you have written, we need to solve
a+b+c+d+e=50
a+b+c+d+e=50
and
a+5b+10c+20d+50e=100
a+5b+10c+20d+50e=100
where a,b,c,d,e∈Na,b,c,d,e∈N. A good way in my opinion is to look at the problem on a case by case basis and solve.
It is better to start from ee and go all the way up to aa looking at all possible options for the values taken by e,d,c,b,ae,d,c,b,a. Further note that 5|a5|a.
A blind upper bound for the values taken by a,b,c,d,ea,b,c,d,e can be easily obtained.
a≤50a≤50, b≤20b≤20, c≤10c≤10, d≤5d≤5 and e≤2e≤2
Now we look at the different cases. There are not a lot of cases and in most of the cases we can easily reject it.
e=2e=2 Not possible
e=1e=1 We need a+b+c+d=49a+b+c+d=49 and a+5b+10c+20d=50a+5b+10c+20d=50 Again not possible
Hence e=0e=0. Hence, 5050 rupee note is not needed.
d=5d=5 Not possible
d=4d=4 We need a+b+c=46a+b+c=46 and a+5b+10c=20a+5b+10c=20. Note that since a,b,c∈Na,b,c∈N we always need to have a+5b+10c≥a+b+ca+5b+10c≥a+b+c. Hence, this case is also ruled out.
d=3d=3 We need a+b+c=47a+b+c=47 and a+5b+10c=40a+5b+10c=40. Same argument as the previous case rules this out.
d=2d=2 We need a+b+c=48a+b+c=48 and a+5b+10c=60a+5b+10c=60. Look at the possible sub cases now. Note that we need a+5b≥a+ba+5b≥a+b and hence we need 60−10c≥48−c60−10c≥48−c and hence we need c≤43c≤43 and hence c∈{0,1}c∈{0,1}
d=2,c=0d=2,c=0 We need a+b=48a+b=48 and a+5b=60a+5b=60. We get a=45a=45 and b=3b=3.
d=2,c=1d=2,c=1 We need a+b=47a+b=47 and a+5b=50a+5b=50. Not possible since a,b∈Na,b∈N
d=1d=1. In this case, we get a+b+c=49a+b+c=49 and a+5b+10c=80a+5b+10c=80. Look at the possible subcases now. Note that we need a+5b≥a+ba+5b≥a+b and hence we need 80−10c≥49−c80−10c≥49−c and hence c≤319c≤319 and hence c∈{0,1,2,3}c∈{0,1,2,3}. Further 4b=(a+5b)−(a+b)=(80−10c)−(49−c)=31−9c4b=(a+5b)−(a+b)=(80−10c)−(49−c)=31−9c. Hence 4|(31−9c)4|(31−9c). Hence the only possibility is when c=3c=3
d=1,c=3d=1,c=3. We get a+b=46a+b=46 and a+5b=50a+5b=50 which gives us a=45a=45 and b=1b=1
d=0d=0. In this case, we need a+b+c=50a+b+c=50 and a+5b+10c=100a+5b+10c=100. Look at the possible subcases now. Note that we need a+5b≥a+ba+5b≥a+b and hence we need 10−10c≥50−c10−10c≥50−c and hence c≤509c≤509 and hence c∈{0,1,2,3,4,5}c∈{0,1,2,3,4,5}. Further 4b=(a+5b)−(a+b)=(100−10c)−(50−c)=50−9c4b=(a+5b)−(a+b)=(100−10c)−(50−c)=50−9c. Hence 4|(50−9c)4|(50−9c). Hence the only possibility is when c=2c=2
d=0,c=2d=0,c=2. We need a+b=48a+b=48 and a+5b=80a+5b=80. Solving this we get a=40,b=8a=40,b=8
Hence, the only possible solutions are