Cryptarithmetic








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Quiz Begins Here

Q #1
⏰ :

Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .

Here OAT is a 3 digit number , BO0T is a 4- digit number . OAT+OAT+OAT=BOOT if t=0 then what will the value of TOO+TOO?

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Discuss
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Explanation: OAT+OAT+OAT = BOOT. As T=0, no carry for A+A+A(3A). Possible values of A and O can be calculated by 1) 3A= O 2) 3A = 10 +O 3) 3A = 20 + O Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of O is less than 7 for equation 3. For Equation 3) Assume value of O is 7. Therefore value of A=9 now carry + O + O + O = BO. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of O as 4 for equation 3 O=4 therefore, A = 8. now carry + O + O + O = BO. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence O=4 A=8 and B=1 now TOO + TOO = TAA 044 + 044 = 088

Q #2
⏰ :

Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .

Here OAT is a 3 digit number , BO0T is a 4- digit number. OAT+OAT+OAT=BOOT if t=0 then what will the value of TO+TO?

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Discuss
+

Explanation: OAT+OAT+OAT = BOOT. As T=0, no carry for A+A+A(3A). Possible values of A and O can be calculated by 1) 3A= O 2) 3A = 10 +O 3) 3A = 20 + O Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of O is less than 7 for equation 3. For Equation 3) Assume value of O is 7. Therefore value of A=9 now carry + O + O + O = BO. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of O as 4 for equation 3 O=4 therefore, A = 8. now carry + O + O + O = BO. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence O=4 A=8 and B=1 now TOO + TOO = TAA

Q #3
⏰ :

Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .

Here OAT is a 3 digit number , BO0T is a 4- digit number . OAT+OAT+OAT=BOOT if t=0 then what will the value of BOO - TOO?

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Discuss
+

Explanation: OAT+OAT+OAT = BOOT. As T=0, no carry for A+A+A(3A). Possible values of A and O can be calculated by 1) 3A= O 2) 3A = 10 +O 3) 3A = 20 + O Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of O is less than 7 for equation 3. For Equation 3) Assume value of O is 7. Therefore value of A=9 now carry + O + O + O = BO. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of O as 4 for equation 3 O=4 therefore, A = 8. now carry + O + O + O = BO. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence O=4 A=8 and B=1 now TOO + TOO = TAA

Q #4
⏰ :

Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .

Here OAT is a 3 digit number , BO0T is a 4- digit number . OAT+OAT+OAT=BOOT if t=0 then what will the value of OA*TO +TOO?

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Discuss
+

Explanation: OAT+OAT+OAT = BOOT. As T=0, no carry for A+A+A(3A). Possible values of A and O can be calculated by 1) 3A= O 2) 3A = 10 +O 3) 3A = 20 + O Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of O is less than 7 for equation 3. For Equation 3) Assume value of O is 7. Therefore value of A=9 now carry + O + O + O = BO. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of O as 4 for equation 3 O=4 therefore, A = 8. now carry + O + O + O = BO. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence O=4 A=8 and B=1 now TOO + TOO = TAA

Q #5
⏰ :

Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .

Here OAT is a 3 digit number , BO0T is a 4- digit number . OAT+OAT+OAT=BOOT if t=0 then what will the value of TOO - OO?

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Discuss
+

Explanation: OAT+OAT+OAT = BOOT. As T=0, no carry for A+A+A(3A). Possible values of A and O can be calculated by 1) 3A= O 2) 3A = 10 +O 3) 3A = 20 + O Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of O is less than 7 for equation 3. For Equation 3) Assume value of O is 7. Therefore value of A=9 now carry + O + O + O = BO. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of O as 4 for equation 3 O=4 therefore, A = 8. now carry + O + O + O = BO. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence O=4 A=8 and B=1 now TOO + TOO = TAA