Sunday Times Teaser 2740 – X Times

by Victor Bryant

Published: 28 March 2015 (link)

					–		X		–
					–		X		–
=	=	=	=	=	=	=	=	=	=
	–		–		–		–		–
	–		–		X		X		–
			–		–		–		–
=	=	=	=	=	=	=	=	=	=
	X		–		–		–		–
=	=	=	=	=	=	=	=	=	=

In this long multiplication sum, I am multiplying a three-figure number by itself. Throughout the workings one particular digit has been replaced by X wherever it occurs: all other digits have been replaced by a dash.

What is the three-figure number being squared?

3 Comments Leave one →

Brian Gladman permalink


# from the middle intermediate sum we can restrict the lower two digits 
# of the number to be squared since (10.x + y).x == 11.x % 100
s_xy = set(xy for xy in range(10, 100) if not ((xy - 11) * (xy // 10)) % 100)

# w is the hundreds digit of the number that is squared 
for w in range(1, 10):
  for xy in s_xy:
    nbr = 100 * w + xy
    # now find the three intermediate products
    hi, mid, lo = (nbr * int(x) for x in str(nbr))
    # and check that they are all in the right range
    if 100 <= hi < 1000 and 1000 <= mid < 10000 and 1000 <= lo < 10000:
      # now check that the highest digit or the result is correct
      if nbr ** 2 // 10000 == (xy // 10):
        print('The number being squared is {}.n'.format(nbr))
        print('        ', ' '.join(x for x in str(nbr)))
        print('     x  ', ' '.join(x for x in str(nbr)))
        print('    ', '==========')
        print('    ', ' '.join(x for x in str(100 * hi)))
        print('    ', ' '.join(x for x in str(10 * mid)))
        print('      ', ' '.join(x for x in str(lo)))
        print('    ', '==========')
        print('    ', ' '.join(x for x in str(nbr ** 2)))
        print('    ', '==========n')

# from the middle intermediate sum we can restrict the lower two digits

# of the number to be squared since (10.x + y).x == 11.x % 100

s_xy = set(xy for xy in range(10, 100) if not ((xy - 11) * (xy // 10)) % 100)

# w is the hundreds digit of the number that is squared

for w in range(1, 10):

for xy in s_xy:

nbr = 100 * w + xy

# now find the three intermediate products

hi, mid, lo = (nbr * int(x) for x in str(nbr))

# and check that they are all in the right range

if 100 <= hi < 1000 and 1000 <= mid < 10000 and 1000 <= lo < 10000:

# now check that the highest digit or the result is correct

if nbr ** 2 // 10000 == (xy // 10):

print('The number being squared is {}.n'.format(nbr))

print(' ', ' '.join(x for x in str(nbr)))

print(' x ', ' '.join(x for x in str(nbr)))

print(' ', '==========')

print(' ', ' '.join(x for x in str(100 * hi)))

print(' ', ' '.join(x for x in str(10 * mid)))

print(' ', ' '.join(x for x in str(lo)))

print(' ', '==========')

print(' ', ' '.join(x for x in str(nbr ** 2)))

print(' ', '==========n')

GeoffR permalink


from itertools import permutations

for p1 in permutations('1234567890', 3):
    a, X, b = p1
    aXb = int(a + X + b)
    # check leading digit of multiplication result
    res = aXb * aXb
    if res //10000 != int(X):continue
    
    # 1st line of multiplication
    line1 = int(a) * aXb * 100
    if X in set(str(line1)):continue
    
    # 2nd line of multiplication
    line2 = int(X) * aXb * 10
    if line2 //100 % 10 != int(X):continue
    if line2//10 % 10 != int(X):continue
    
    # 3rd line of multiplication
    line3 = int(b) * aXb
    if X in set(str(line3)):continue
    print(f"The three-figure number being squared is {aXb}.")

from itertools import permutations

for p1 in permutations('1234567890', 3):

a, X, b = p1

aXb = int(a + X + b)

# check leading digit of multiplication result

res = aXb * aXb

if res //10000 != int(X):continue

# 1st line of multiplication

line1 = int(a) * aXb * 100

if X in set(str(line1)):continue

# 2nd line of multiplication

line2 = int(X) * aXb * 10

if line2 //100 % 10 != int(X):continue

if line2//10 % 10 != int(X):continue

# 3rd line of multiplication

line3 = int(b) * aXb

if X in set(str(line3)):continue

print(f"The three-figure number being squared is {aXb}.")