+1833 $$The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of the three integers?$$
The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of the three integers?
+33615 This must be of the form a/60 + b/60 + c/60 = 47/60
But since it must also be of the form 1/p + 1/q + 1/r = 47/60 then a, b and c must be factors of 60 (so that each term on the LHS cancels to leave a 1 in the numerator).
$${factor}{\left({\mathtt{60}}\right)} = {{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}$$
Clearly the 3 consecutive numbers are 3, 4 and 5, so their sum is 12.
Check
$${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\frac{{\mathtt{47}}}{{\mathtt{60}}}} = {\mathtt{0.783\: \!333\: \!333\: \!333\: \!333\: \!3}}$$
.
+33615 This must be of the form a/60 + b/60 + c/60 = 47/60
But since it must also be of the form 1/p + 1/q + 1/r = 47/60 then a, b and c must be factors of 60 (so that each term on the LHS cancels to leave a 1 in the numerator).
$${factor}{\left({\mathtt{60}}\right)} = {{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}$$
Clearly the 3 consecutive numbers are 3, 4 and 5, so their sum is 12.
Check
$${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\frac{{\mathtt{47}}}{{\mathtt{60}}}} = {\mathtt{0.783\: \!333\: \!333\: \!333\: \!333\: \!3}}$$
.
