Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .
Explanation:
Two digit maximum addition is (9+9) 18,
Since carry is there as letter B. Therefore B has to be 1(B =1).
Now T+B has to be greater than 10(it can be 10, 11, 12,...18).Since B=1
T+B will not exceed greater than 11 including carry on T (0 or 1).
Therefore O=0
Now to get carry as 1, T must be either 9 or 8, but carry may be there on T (0 or 1).
Now E+O (Since O=0) will come E, but we want it different.
Therefore on E there is carry as 1.
Now N+R must be greater than 10, Therefore (carry (0 or 1) + N+R) =E
Carry+E+1+R=E. Therefore R=8
And T must be then 9.
D+E greater than 10
Therefore D=7 and E=5, and N=E+1=6, and y=2
Therefore B=1, O=0, N=6, E=5, Y=2.
Therefore, B+O+N+E+Y = 1+0+6+5+2 = 14
Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .
Explanation:
Two digit maximum addition is (9+9) 18,
Since carry is there as letter B. Therefore B has to be 1(B =1).
Now T+B has to be greater than 10(it can be 10, 11, 12,...18).Since B=1
T+B will not exceed greater than 11 including carry on T (0 or 1).
Therefore O=0
Now to get carry as 1, T must be either 9 or 8, but carry may be there on T (0 or 1).
Now E+O (Since O=0) will come E, but we want it different.
Therefore on E there is carry as 1.
Now N+R must be greater than 10, Therefore (carry (0 or 1) + N+R) =E
Carry+E+1+R=E. Therefore R=8
And T must be then 9.
D+E greater than 10
Therefore D=7 and E=5, and N=E+1=6, and y=2
Therefore B=1, O=0, N=6, E=5, Y=2.
Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .
Explanation:
Two digit maximum addition is (9+9) 18,
Since carry is there as letter B. Therefore B has to be 1(B =1).
Now T+B has to be greater than 10(it can be 10, 11, 12,...18).Since B=1
T+B will not exceed greater than 11 including carry on T (0 or 1).
Therefore O=0
Now to get carry as 1, T must be either 9 or 8, but carry may be there on T (0 or 1).
Now E+O (Since O=0) will come E, but we want it different.
Therefore on E there is carry as 1.
Now N+R must be greater than 10, Therefore (carry (0 or 1) + N+R) =E
Carry+E+1+R=E. Therefore R=8
And T must be then 9.
D+E greater than 10
Therefore D=7 and E=5, and N=E+1=6, and y=2
Therefore B=1, O=0, N=6, E=5, Y=2.
Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .
Explanation:
Two digit maximum addition is (9+9) 18,
Since carry is there as letter B. Therefore B has to be 1(B =1).
Now T+B has to be greater than 10(it can be 10, 11, 12,...18).Since B=1
T+B will not exceed greater than 11 including carry on T (0 or 1).
Therefore O=0
Now to get carry as 1, T must be either 9 or 8, but carry may be there on T (0 or 1).
Now E+O (Since O=0) will come E, but we want it different.
Therefore on E there is carry as 1.
Now N+R must be greater than 10, Therefore (carry (0 or 1) + N+R) =E
Carry+E+1+R=E. Therefore R=8
And T must be then 9.
D+E greater than 10
Therefore D=7 and E=5, and N=E+1=6, and y=2
Therefore B=1, O=0, N=6, E=5, Y=2.
Therefore, B+O+N+E+Y = 1+0+6+5+2 = 14
Directions: In this Question , Each Alphabet used is for a distinct digit & the equation given is mathematically true .
Explanation:
Two digit maximum addition is (9+9) 18,
Since carry is there as letter B. Therefore B has to be 1(B =1).
Now T+B has to be greater than 10(it can be 10, 11, 12,...18).Since B=1
T+B will not exceed greater than 11 including carry on T (0 or 1).
Therefore O=0
Now to get carry as 1, T must be either 9 or 8, but carry may be there on T (0 or 1).
Now E+O (Since O=0) will come E, but we want it different.
Therefore on E there is carry as 1.
Now N+R must be greater than 10, Therefore (carry (0 or 1) + N+R) =E
Carry+E+1+R=E. Therefore R=8
And T must be then 9.
D+E greater than 10
Therefore D=7 and E=5, and N=E+1=6, and y=2
Therefore B=1, O=0, N=6, E=5, Y=2.
Therefore, B+O+N+E+Y = 1+0+6+5+2 = 14