Respuesta :
To answer this question, you need to make a list of the condition that has to be met.
To made it easier, let say for number 54,321 = ed,cba
a=1 (oneth)
b=2(tenth)
c=3(hundreth)
d=4(thousanth)
e=5(tenthousanth)
Condition that doesn't refer each other should be put on top
1. The number has seven digits.
2. None of the digits are zero
3. The number is between 1,000 and 10,000 (mean e=1 or e=0. notice that e can be 0 because it the last digit so it if it 0 then it won't be written)
Then the condition that refers each other should be:
4. The digit in the tenths place is %150 of the digit in the thousands place.(b= 1.5d)
5. The digit in the hundredths place is 6/5 of the sum of the digits in the tenths place and the digit in the thousands place.[c= 6/5(b+d)]
6. The sum of all the digits in the number is 17. (a+b+c+d+e=17)
Inserting number 4 into number 5 would result
c=6/5(1.5d+d)
c= 6/5(2.5d)
7. c=3d
From here you can convert b into d(equation 4) and convert c into d(equation 7). If you insert both into equation 6 it would be:
a+(1.5d)+(3d)+d+e= 17
a+ 5.5d+e=17
From equation 4 (b= 1.5d) it is clear that d cannot be an odd number because the result will have 0.5(digit shouldn't have fraction).
From this point, it is clear that d should be 2 because 3 would result in 0.5 number and >4 will exceed 17 and 1 are odd number. (remember no 0 allowed).
Put d=2 into equation 7 will result:
a+ 5.5(2)+e=17
a+e= 17-11
a+e=6
since e possibility only 0 and 1, then a possibility is only 5 or 6. The list would be:
12,635
2,636
To made it easier, let say for number 54,321 = ed,cba
a=1 (oneth)
b=2(tenth)
c=3(hundreth)
d=4(thousanth)
e=5(tenthousanth)
Condition that doesn't refer each other should be put on top
1. The number has seven digits.
2. None of the digits are zero
3. The number is between 1,000 and 10,000 (mean e=1 or e=0. notice that e can be 0 because it the last digit so it if it 0 then it won't be written)
Then the condition that refers each other should be:
4. The digit in the tenths place is %150 of the digit in the thousands place.(b= 1.5d)
5. The digit in the hundredths place is 6/5 of the sum of the digits in the tenths place and the digit in the thousands place.[c= 6/5(b+d)]
6. The sum of all the digits in the number is 17. (a+b+c+d+e=17)
Inserting number 4 into number 5 would result
c=6/5(1.5d+d)
c= 6/5(2.5d)
7. c=3d
From here you can convert b into d(equation 4) and convert c into d(equation 7). If you insert both into equation 6 it would be:
a+(1.5d)+(3d)+d+e= 17
a+ 5.5d+e=17
From equation 4 (b= 1.5d) it is clear that d cannot be an odd number because the result will have 0.5(digit shouldn't have fraction).
From this point, it is clear that d should be 2 because 3 would result in 0.5 number and >4 will exceed 17 and 1 are odd number. (remember no 0 allowed).
Put d=2 into equation 7 will result:
a+ 5.5(2)+e=17
a+e= 17-11
a+e=6
since e possibility only 0 and 1, then a possibility is only 5 or 6. The list would be:
12,635
2,636
