Guest Posted February 14, 2010 Report Share Posted February 14, 2010 (edited) Substitute each of the capital letters by a different base ten digit from 0 to 9 to satisfy this set of alphametic relationships. None of the numbers can contain any leading zero. THOMAS+MALTHUS+1766+1834+RR=RRRRRRR, and: MOST is divisible by 23 Edited February 14, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 14, 2010 Report Share Posted February 14, 2010 T=8 H=4 O=0 M=1 A=3 S=5 L=7 U=6 R=2 yielding: 840135+1378465+22+3600=2222222 great alphametic. took a lot of thinking. Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted February 18, 2010 Report Share Posted February 18, 2010 I agree with the magic man. His answer and what a fun challenge this one is. Deserves more than one twenty couple click thrus in my opinion hence the bump. Gonna post my methodology but strongly suggest giving it a go. Know KS usually doesnt post answers and in no way would I want to suggest he should or do anything that might otherwise distract him from posting more fabulous teasers so here goes (besides, I'm a little proud I actually got one) MALTHUS THOMAS + 3600 ======= RRRRR00 S+S=10 so S=5 (consequently U+A=9 and H+M+7= R or R+10 or R+20) 11 MALTHU5 THOMA5 + 3600 ======= RRRRR00 Subtract RR from both sides of the equation: 11 MALTHU5 THOMA5 + 3600 ======= RRRRR00 M+1=R (A+T<18 and if A+T<10 then M=R) H+M+7= R or R+10 or R+20 if H+M+7=R then M+1=H+M+7 or H=-6 so no if H+M+7=R+20 then M+1=H+M-13 or H=12 so no if H+M+7=R+10 then M+1=H+M-3 or H=4 1 111 MALT4U5 T4OMA5 + 3600 ======= RRRRR00 Solve for R in the left most column of the equation and in the third column from the right and set them equal to one another: M,O,S,T are unique MO5T/23 is an integer M=(1,2,6,7,8) (M cannot equal 0 (condition of the OP),3(if M=3 then R=4 but H=4),4(H=4),5(S=5) or 9(the sum would have too many digits)) I used a spreadsheet and with the above restrictions, narrowed down MOST=(1058,1357,7153,7659,7958,8257) Limit the possibilities of MOST: 1 111 MALT4U5 T4OMA5 + 3600 ======= RRRRR00 A+T(+1 if L+4>9)=R+10 T+O+4= R or R+10 or R+20 if T+O+4=R then A+T-(10 or 9)=T+O+4 or A-(14 or 13)=O so no if T+O+4=R+20 and R>1 (since M+1=R) then T+O>17 or T=O=9 so no if T+O+4=R+10 then T+O+4=A+T(+1) or O+(3 or 4)=A so: (O,A)=(0,3)(2,6)(3,6)(3,7)(6,9) now: if O=6 then the fourth column of the equation yields T=R so no if O=3 then MOST=1357 (see STEP 3) and T=7 and plugging in O=3 and T=7 in the fourth column yields R=4 but H=4 so no if O=2 then MOST=8257 and M=8 R=9 so looking at the third column from the left L would have to equal 5 but S=5 so no. so O=0 and A=3 and MOST=1058 and ...plug and chug Am sure there are other methods and undoubtedly a more elegant solution but that's what I gotsk. Much fun KS! Solve the second column from the left and the fourth column simultaneously: Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 19, 2010 Report Share Posted February 19, 2010 looks complicated plainglazed. i have no idea what any of you guys were doing. Quote Link to comment Share on other sites More sharing options...

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Substitute each of the capital letters by a different base ten digit from 0 to 9 to satisfy this set of alphametic relationships. None of the numbers can contain any leading zero.

THOMAS+MALTHUS+1766+1834+RR=RRRRRRR, and:MOSTis divisible by 23Edited by K Sengupta## Link to comment

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