The ten numbers in the star total 75.

Each number appears in two "lines".

And there are five lines, making a total of 150. Which implies the total for each line is 30.

This means that R+I=24 and therefore R=11 and I = 13 (or vice versa).

If R=11 then N=12, H =10 and C=11 which is impossible.

If R=13 then N = 10; H = 12 and C= 9, which is correct.

*Alternatively:*

once you know that each line adds up to 30 we can set up the following equations:

$(1) 4+N+R+3=30 ⇒N+R=23$

$(2) 1+R+I+5=30 ⇒R+I=24$

$(3) 3+I+C+7=30 ⇒I+C=20$

$(4) 4+H+C+5=30 ⇒H+C=21$

$(5) 1+N+H+7=30 ⇒N+H=22$

Then we are going to use a trick to “isolate R” by alternately adding and subtracting equations to get what we want, so consider

$(N+R)-(N+H)+(H+C)-(I+C)+(R+I)=23-22+21-20+24$

$⇒2R=26 ⇒R=13.$

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*

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