Question

how to get 50 notes from 100 rupees. 2rupees note not allowed. I want 3 possibilities.

Answer 1

Answer:

As you have written, we need to solve

a+b+c+d+e=50

and

a+5b+10c+20d+50e=100

where a,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 e and go all the way up to a looking at all possible options for the values taken by e,d,c,b,a. Further note that 5|a.

A blind upper bound for the values taken by a,b,c,d,e can be easily obtained.

a≤50, b≤20, c≤10, d≤5 and e≤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=2 Not possible

e=1 We need a+b+c+d=49 and a+5b+10c+20d=50 Again not possible

Hence e=0. Hence, 50 rupee note is not needed.

d=5 Not possible

d=4 We need a+b+c=46 and a+5b+10c=20. Note that since a,b,c∈N we always need to have a+5b+10c≥a+b+c. Hence, this case is also ruled out.

d=3 We need a+b+c=47 and a+5b+10c=40. Same argument as the previous case rules this out.

d=2 We need a+b+c=48 and a+5b+10c=60. Look at the possible sub cases now. Note that we need a+5b≥a+b and hence we need 60−10c≥48−c and hence we need c≤43 and hence c∈{0,1}

d=2,c=0 We need a+b=48 and a+5b=60. We get a=45 and b=3.

d=2,c=1 We need a+b=47 and a+5b=50. Not possible since a,b∈N

d=1. In this case, we get a+b+c=49 and a+5b+10c=80. Look at the possible subcases now. Note that we need a+5b≥a+b and hence we need 80−10c≥49−c and hence c≤319 and hence c∈{0,1,2,3}. Further 4b=(a+5b)−(a+b)=(80−10c)−(49−c)=31−9c. Hence 4|(31−9c). Hence the only possibility is when c=3

d=1,c=3. We get a+b=46 and a+5b=50 which gives us a=45 and b=1

d=0. In this case, we need a+b+c=50 and a+5b+10c=100. Look at the possible subcases now. Note that we need a+5b≥a+b and hence we need 10−10c≥50−c and hence c≤509 and hence c∈{0,1,2,3,4,5}. Further 4b=(a+5b)−(a+b)=(100−10c)−(50−c)=50−9c. Hence 4|(50−9c). Hence the only possibility is when c=2

d=0,c=2. We need a+b=48 and a+5b=80. Solving this we get a=40,b=8

Hence, the only possible solutions are

{45,3,0,2,0}

{45,1,3,1,0}

{40,8,2,0,0}